2.1. Stronger topologies on $\mathcal{P}(E)$

As a last preparation, we discuss a well-known class of topologies on subsets of $\mathcal{P}(E)$ with which we will work frequently. Given a continuous function $\psi \colon E \rightarrow {\mathbb R}_+ \,:\!= [0,\infty)$ , define

\[\mathcal{P}_\psi(E) = \bigg\{\mu \in \mathcal{P}(E) \colon \int_E\psi\,{\mathrm{d}} \mu < \infty\bigg\}.\]

Endow $\mathcal{P}_\psi(E)$ with the (Polish) topology generated by the maps $\nu \mapsto \int_Ef\,{\mathrm{d}} \nu$ , where $f \colon E \rightarrow {\mathbb R}$ is continuous and $|\,f| \le 1+\psi$ ; we call this the $\psi$ -weak topology. A useful fact about this topology is that a set $M \subset \mathcal{P}_\psi(E)$ is pre-compact if and only if for every $\epsilon \gt 0$ there exists a compact set $K \subset E$ such that

\[\sup_{\mu \in M}\int_{K^c}\psi\,{\mathrm{d}} \mu \le \epsilon.\]

This is easily proved directly using Prokhorov’s theorem, or refer to [28, Corollary A.47]. It is worth noting that if d is a compatible metric on E and $\psi(x)=d^p(x,x_0)$ for some fixed $x_0 \in E$ and $p \ge 1$ , then the $\psi$ -weak topology is simply the p-Wasserstein topology associated with the metric d [Reference Villani52, Theorem 7.12].

2.2. An extension of Theorem 1.1

In this section we state and prove a useful generalization of Theorem 1.1 for stronger topologies and unbounded functions, taking advantage of the preparations of the previous sections. At all times in this section, the standing assumptions on $(\alpha,\rho)$ (stated early in Section 2) are in force. Part of Theorem 2.3 below requires the assumption that the sub-level sets of $\alpha$ are pre-compact in $\mathcal{P}_\psi(E)$ , and this rather opaque assumption will be explored in more detail in Section 3.1.

Theorem 2.3. Let $\psi \colon E \rightarrow {\mathbb R}_+$ be continuous. If $F \colon \mathcal{P}_\psi(E) \rightarrow {\mathbb R} \cup \{\infty\}$ is lower semicontinuous (with respect to the $\psi$ -weak topology) and bounded from below, then

\[\liminf_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF \circ L_n) \ge \sup_{\nu \in \mathcal{P}_\psi(E)}(F(\nu) - \alpha(\nu)).\]

Suppose also that the sub-level sets of $\alpha$ are pre-compact subsets of $\mathcal{P}_\psi(E)$ . If $F \colon \mathcal{P}_\psi(E) \rightarrow {\mathbb R} \cup \{-\infty\}$ is upper semicontinuous and bounded from above, then

\[\limsup_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF \circ L_n) \le \sup_{\nu \in \mathcal{P}_\psi(E)}(F(\nu) - \alpha(\nu)).\]

Proof of lower bound. Let us first prove the lower bound. It is immediate from the definition that $n^{-1}\alpha_n(\nu^n) = \alpha(\nu)$ for each $\nu \in \mathcal{P}(E)$ , recalling that $\nu^n$ denotes the n-fold product measure. Thus

(2.2) \begin{align}\dfrac{1}{n}\rho_n(nF(L_n)) &= \sup_{\nu \in \mathcal{P}(E^n)}\bigg\{\int_{E^n} F \circ L_n \,{\mathrm{d}} \nu - \dfrac{1}{n}\alpha_n(\nu)\bigg\} \notag \\* &\ge \sup_{\nu \in \mathcal{P}(E)}\bigg\{\int_{E^n} F \circ L_n \,{\mathrm{d}} \nu^n - \dfrac{1}{n}\alpha_n(\nu^n)\bigg\} \notag \\* &= \sup_{\nu \in \mathcal{P}(E)}\bigg\{\int_{E^n} F \circ L_n \,{\mathrm{d}} \nu^n - \alpha(\nu)\bigg\}.\end{align}

For $\nu \in \mathcal{P}(E)$ , the law of large numbers (see [Reference Dudley20, Theorem 11.4.1]) implies $\nu^n \circ L_n^{-1} \rightarrow \delta_\nu$ weakly, i.e. in $\mathcal{P}(\mathcal{P}(E))$ . For $\nu \in \mathcal{P}_{\psi}(E)$ , the convergence takes place in $\mathcal{P}(\mathcal{P}_{\psi}(E))$ . Lower semicontinuity of F on $\mathcal{P}_\psi(E)$ then implies (e.g. by [Reference Dupuis and Ellis22, Theorem A.3.12]), for each $\nu \in \mathcal{P}_\psi(E)$ ,

\begin{align*}\liminf_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF(L_n)) &\ge \liminf_{n\rightarrow\infty}\int_{E^n} F \circ L_n \,{\mathrm{d}} \nu^n - \alpha(\nu) \\* &\ge F(\nu) - \alpha(\nu).\end{align*}

Take the supremum over $\nu$ to complete the proof of the lower bound.

Proof of upper bound, F bounded. The upper bound is more involved. First we prove it in four steps under the assumption that F is bounded.

Step 1. First we simplify the expression somewhat. For each $\nu \in \mathcal{P}(E^n)$ the definition of $\alpha_n$ and convexity of $\alpha$ imply

\begin{align*}\dfrac{1}{n}\alpha_n(\nu) &= \dfrac{1}{n}\sum_{k=1}^n\int_{E^n}\alpha(\nu_{k-1,k}(x_1,\ldots,x_{k-1}) )\nu({\mathrm{d}} x_1,\ldots,{\mathrm{d}} x_n) \\* &\ge \int_{E^n}\alpha\Bigg(\dfrac{1}{n}\sum_{k=1}^n\nu_{k-1,k}(x_1,\ldots,x_{k-1})\Bigg)\nu({\mathrm{d}} x_1,\ldots,{\mathrm{d}} x_n).\end{align*}

Combine this with (2.2) to get

(2.3) \begin{equation}\dfrac{1}{n}\rho_n(nF(L_n)) \le \sup_{\nu \in \mathcal{P}(E^n)}\int_{E^n}\Bigg[F(L_n) - \alpha\Bigg(\dfrac{1}{n}\sum_{k=1}^n\nu_{k-1,k}\Bigg)\Bigg]\,{\mathrm{d}} \nu.\end{equation}

Now choose arbitrarily some $\mu_f$ such that $\alpha(\mu_f) \lt \infty$ . The choice $\nu = \mu_f^n$ and boundedness of F show that the supremum in (2.3) is bounded below by $-\|F\|_{\infty} - \alpha(\mu_f)$ , where $\|F\|_\infty \,:\!= \sup_{\nu \in \mathcal{P}_\psi(E)}|F(\nu)|$ . For each n, choose $\nu^{(n)} \in \mathcal{P}(E^n)$ attaining the supremum in (2.3) to within $1/n$ . Then

(2.4) \begin{equation}\int_{E^n}\alpha\Bigg(\dfrac{1}{n}\sum_{k=1}^n\nu^{(n)}_{k-1,k}\Bigg)\,{\mathrm{d}} \nu^{(n)} \le 2\|F\|_{\infty} + \alpha(\mu_f) + \dfrac{1}{n}.\end{equation}

It is convenient to switch now to a probabilistic notation. On some sufficiently rich probability space, find an $E^n$ -valued random variable $(Y^n_1,\ldots,Y^n_n)$ with law $\nu^{(n)}$ . Define the random measures

\[S_n \,:\!= \dfrac{1}{n}\sum_{k=1}^n\nu^{(n)}_{k-1,k}(Y^n_1,\ldots,Y^n_{k-1}), \quad \skew3\widetilde{S}_n \,:\!= \dfrac{1}{n}\sum_{k=1}^n\delta_{Y^n_k}.\]

Use (2.3) and unwrap the definitions to find

(2.5) \begin{equation}\dfrac{1}{n}\rho_n(nF(L_n)) \le {\mathbb E}[F(\skew3\widetilde{S}_n) - \alpha(S_n)] + 1/n.\end{equation}

Moreover, (2.4) implies

(2.6) \begin{equation}\sup_n{\mathbb E}[\alpha(S_n)] \le 2\|F\|_\infty + \alpha(\mu_f) + 1 < \infty.\end{equation}

Step 2. We next show that the sequence $(S_n,\skew3\widetilde{S}_n)$ is tight, viewed as $\mathcal{P}_\psi(E)\times\mathcal{P}_\psi(E)$ -valued random variables. Here we use the assumption that the sub-level sets of $\alpha$ are $\psi$ -weakly compact subsets of $\mathcal{P}_\psi(E)$ . It then follows from (2.6) that $(S_n)$ is tight (see e.g. [Reference Dupuis and Ellis22, Theorem A.3.17]).

To see that the pair $(S_n,\skew3\widetilde{S}_n)$ is tight, it remains to check that $(\skew3\widetilde{S}_n)_n$ is tight. To this end, we first notice that $S_n$ and $\skew3\widetilde{S}_n$ have the same mean measure for each n, in the sense that for every $f \in B(E)$ we have

(2.7) \begin{align} {\mathbb E}\bigg[\int_Ef\,{\mathrm{d}} S_n\bigg] &= {\mathbb E}\Bigg[\dfrac{1}{n}\sum_{k=1}^n{\mathbb E} [ f(Y^n_k) \mid Y^n_1,\ldots,Y^n_{k-1}]\Bigg] \notag \\* & = {\mathbb E}\Bigg[\dfrac{1}{n}\sum_{k=1}^nf(Y^n_k)\Bigg] \notag \\ &= {\mathbb E}\bigg[\int_Ef\,{\mathrm{d}}\skew3\widetilde{S}_n\bigg].\end{align}

To prove $(\skew3\widetilde{S}_n)$ is tight, it suffices (by Prokhorov’s theorem) to show that for all $\epsilon \gt 0$ there exists a $\psi$ -weakly compact set $K \subset \mathcal{P}_\psi(E)$ such that $P(\skew3\widetilde{S}_n \notin K) \le \epsilon$ . We will look for K of the form

\[K = \cap_{k=1}^\infty\bigg\{\nu \colon \int_{C_k^c}\psi\,{\mathrm{d}} \nu \le 1/k\bigg\},\]

where $(C_k)_{k=1}^\infty$ is a sequence of compact subsets of E to be specified later; indeed, sets K of this form are pre-compact in $\mathcal{P}_\psi(E)$ according to a form of Prokhorov’s theorem discussed in Section 2.1 (see also [Reference Föllmer and Schied28, Corollary A.47]). For such a set K, use Markov’s inequality and (2.7) to compute

(2.8) \begin{align}P (\skew3\widetilde{S}_n \notin K ) &\le \sum_{k=1}^\infty P\bigg(\int_{C_k^c}\psi\,{\mathrm{d}}\skew3\widetilde{S}_n> 1/k\bigg) \notag \\*&\le \sum_{k=1}^\infty k\,{\mathbb E}\int_{C_k^c}\psi\,{\mathrm{d}}\skew3\widetilde{S}_n \notag \\*& = \sum_{k=1}^\infty k\,{\mathbb E}\int_{C_k^c}\psi\,{\mathrm{d}} S_n.\end{align}

By a form of Jensen’s inequality (see Lemma B.2),

\[\sup_n\alpha({\mathbb E} S_n) \le \sup_n{\mathbb E}[\alpha(S_n)] < \infty,\]

where ${\mathbb E} S_n$ is the probability measure on E defined by $({\mathbb E} S_n)(A) = {\mathbb E}[S_n(A)]$ . Hence, the sequence $({\mathbb E} S_n)$ is pre-compact in $\mathcal{P}_\psi(E)$ , thanks to the assumption that sub-level sets of $\alpha$ are pre-compact subsets of $\mathcal{P}_\psi(E)$ . It follows that for every $\epsilon \gt 0$ there exists a compact set $C \subset E$ such that $\sup_n{\mathbb E}\int_{C^c}\psi\,{\mathrm{d}} S_n \le \epsilon$ . With this in mind, we may choose $C_k$ to make (2.8) arbitrarily small, uniformly in n. This shows that $(\skew3\widetilde{S}_n)$ is tight, completing Step 2.

Step 3. We next show that every limit in distribution of $(S_n,\skew3\widetilde{S}_n)$ is concentrated on the diagonal $\{(\nu,\nu) \colon \nu \in \mathcal{P}_\psi(E)\}$ . By definition of $\nu^{(n)}_{k-1,k}$ , we have

\[{\mathbb E}\bigg[ f(Y^n_k) - \int_E f\,{\mathrm{d}} \nu^{(n)}_{k-1,k}(Y^n_1,\ldots,Y^n_{k-1})\mid Y^n_1,\ldots,Y^n_{k-1}\bigg] = 0 \quad \text{for } k=1,\ldots,n\]

for every $f \in B(E)$ . That is, the terms inside the expectation form a martingale difference sequence. Thus, for $f \in B(E)$ , we have

(2.9) \begin{align}{\mathbb E}\bigg[\bigg(\int_E f\,{\mathrm{d}} S_n - \int_E f\,{\mathrm{d}}\skew3\widetilde{S}_n\bigg)^2 \bigg] &= {\mathbb E}\Bigg[\Bigg(\dfrac{1}{n}\sum_{k=1}^n\bigg( f(Y^n_k) - \int_E f\,{\mathrm{d}} \nu^{(n)}_{k-1,k}(Y^n_1,\ldots,Y^n_{k-1})\bigg)\Bigg)^2\Bigg] \notag \\* &= \dfrac{1}{n^2}\sum_{k=1}^n{\mathbb E}\bigg[\bigg( f(Y^n_k) - \int_E f\,{\mathrm{d}} \nu^{(n)}_{k-1,k}(Y^n_1,\ldots,Y^n_{k-1})\bigg)^2\bigg] \notag \\* &\le 2\|\,f\|_{\infty}^2/n,\end{align}

where $\|\,f\|_\infty \,:\!= \sup_{x \in E}|\,f(x)|$ . It is straightforward to check that (2.9) implies that every weak limit of $(S_n,\skew3\widetilde{S}_n)$ is concentrated on (i.e. almost surely belongs to) the diagonal $\{(\nu,\nu) \colon \nu \in \mathcal{P}(E)\}$ (cf. [Reference Dupuis and Ellis22, Lemma 2.5.1(b)]). Indeed, if $(S,\skew3\widetilde{S})$ is some $\mathcal{P}_\psi(E) \times \mathcal{P}_\psi(E)$ -valued random variable such that $(S_{n_k},\skew3\widetilde{S}_{n_k})$ converges in law to $(S,\skew3\widetilde{S})$ , then (2.9) implies

\begin{equation*}{\mathbb E}\bigg[\bigg(\int_E f\,{\mathrm{d}} S - \int_E f\,{\mathrm{d}}\skew3\widetilde{S}\bigg)^2\bigg] = 0,\end{equation*}

for each $f \in C_b(E)$ , by continuity of the map

\[ \mathcal{P}_\psi(E) \times \mathcal{P}_\psi(E) \ni (\nu,\widetilde{\nu}) \mapsto \bigg(\int_E f\,{\mathrm{d}} \nu - \int_E f\,{\mathrm{d}}\widetilde{\nu}\bigg)^2.\]

Hence, $\int_E f\,{\mathrm{d}} S = \int_E f\,{\mathrm{d}}\skew3\widetilde{S}$ a.s. for each $f \in C_b(E)$ , and arguing with a countable separating family from $C_b(E)$ (see e.g. [Reference Parthasarathy45, Theorem 6.6]) allows us to deduce that $S=\skew3\widetilde{S}$ a.s.

Step 4. We can now complete the proof of the upper bound. With Step 3 in mind, fix a subsequence and a $\mathcal{P}_\psi(E)$ -valued random variable $\eta$ such that $(S_n,\skew3\widetilde{S}_n) \rightarrow (\eta,\eta)$ in distribution (where we relabeled the subsequence). Recall that $\alpha$ is bounded from below and $\psi$ -weakly lower semicontinuous, whereas F is upper semicontinuous and bounded. Returning to (2.5), we conclude now that

\begin{align*}\limsup_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF(L_n)) &\le \limsup_{n\rightarrow\infty}{\mathbb E} [F(\skew3\widetilde{S}_n) - \alpha(S_n)] \\* &\le {\mathbb E}[F(\eta) - \alpha(\eta)] \\* &\le \sup_{\nu \in \mathcal{P}_\psi(E)} \{F(\nu) - \alpha(\nu)\}.\end{align*}

Of course, we abused notation by relabeling the subsequences, but we have argued that for every subsequence there exists a further subsequence for which this bound holds, which proves the upper bound for F bounded.

Proof of upper bound, unbounded F. With the proof complete for bounded F, we now remove the boundedness assumption using a natural truncation procedure. Let $F \colon \mathcal{P}(E) \rightarrow E \cup \{-\infty\}$ be upper semicontinuous and bounded from above. For $m \gt 0$ let $F_m \,:\!= F \vee ({-}m)$ . Since $F_m$ is bounded and upper semicontinuous, the previous step yields

\[\limsup_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF_m(L_n)) \le \sup_{\nu \in \mathcal{P}_\psi(E)} \{F_m(\nu) - \alpha(\nu)\} =\!:\, S_m,\]

for each $m \gt 0$ . Since $F_m \ge F$ , we have

\[\rho_n(nF_m(L_n)) \ge \rho_n(nF(L_n))\]

for each m, and it remains only to show that

(2.10) \begin{equation}\lim_{m \rightarrow \infty}S_m = \sup_{\nu \in \mathcal{P}_\psi(E)}\{F(\nu) - \alpha(\nu)\} =\!:\, S.\end{equation}

Clearly $S_m \ge S$ , since $F_m \ge F$ . Note that $S \lt \infty$ , as F and $\alpha$ are bounded from above and from below, respectively. If $S = -\infty$ , then $F(\nu) = -\infty$ whenever $\alpha(\nu) \lt \infty$ , and we conclude that, as $m\rightarrow\infty$ ,

\[S_m \le -m - \inf_{\nu \in \mathcal{P}(E)}\alpha(\nu) \ \rightarrow \ -\infty = S.\]

Now suppose instead that S is finite. Fix $\epsilon \gt 0$ . For each $m \gt 0$ , find $\nu_m \in \mathcal{P}(E)$ such that

(2.11) \begin{equation}F_m(\nu_m) - \alpha(\nu_m) + \epsilon \ge S_m \ge S.\end{equation}

Since F is bounded from above and $S \gt -\infty$ , it follows that $\sup_m\alpha(\nu_m) \lt\infty$ . The sub-level sets of $\alpha$ are $\psi$ -weakly compact, and thus the sequence $(\nu_m)$ has a limit point (in $\mathcal{P}_\psi(E)$ ). Let $\nu_\infty$ denote any limit point, and suppose $\nu_{m_k} \rightarrow \nu_\infty$ . Note that $\inf_m F_m(\nu_m) \gt -\infty$ in light of (2.11), because $\alpha$ is bounded from below. Hence, for all sufficiently large m, we have $F_m(\nu_m)=F(\nu_m)$ . Thus

\begin{equation*}\limsup_{k\rightarrow\infty} \{F_{m_k}(\nu_{m_k}) - \alpha(\nu_{m_k})\} \le F(\nu_\infty) - \alpha(\nu_\infty)\le S,\end{equation*}

where the second inequality follows from upper semicontinuity of F and lower semicontinuity of $\alpha$ . This holds for any limit point of the pre-compact sequence $(\nu_m)$ , and it follows from (2.11) that

\[S \le \limsup_{m\rightarrow\infty}S_m \le \limsup_{m\rightarrow\infty}\{F_m(\nu_m) - \alpha(\nu_m)\} + \epsilon \le S + \epsilon.\]

Since $\epsilon \gt 0$ was arbitrary, this proves (2.10).

2.3. Contraction principles and an abstract form of Cramér’s theorem

Viewing Theorem 2.3 as an abstract form of Sanov’s theorem, we may derive from it a form of Cramér’s theorem. The key tool is an analog of the contraction principle from classical large deviations (cf. [Reference Dembo and Zeitouni16, Theorem 4.2.1]). In its simplest form, if $\varphi \colon \mathcal{P}(E) \rightarrow E'$ is continuous for some topological space E’, then for $F \in C_b(E')$ we have from Theorem 1.1

\begin{equation*}\lim_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF \circ \varphi \circ L_n) = \sup_{\nu \in \mathcal{P}(E)} (F(\varphi(\nu)) - \alpha(\nu)) = \sup_{x \in E'}(F(x) - \alpha_\varphi(x)),\end{equation*}

where we define $\alpha_\varphi \colon E' \rightarrow ({-}\infty,\infty]$ by

\[\alpha_\varphi(x) \,:\!= \inf \{\alpha(\nu) \colon \nu \in \mathcal{P}(E), \ \varphi(\nu) = x \}.\]

This line of reasoning leads to the following extension of Cramér’s theorem.

Theorem 2.4. Let $(E,\|\cdot\|)$ be a separable Banach space with continuous dual $E^*$ . Define $\Lambda^* \colon E \rightarrow {\mathbb R} \cup \{\infty\}$ by

\begin{equation*}\Lambda^*(x) = \sup_{x^* \in E^*} (\langle x^*,x\rangle - \rho(x^*)).\end{equation*}

Define $S_n \colon E^n \rightarrow E$ by

\[S_n(x_1,\ldots,x_n) = \dfrac{1}{n}\sum_{i=1}^nx_i.\]

If $F \colon E \rightarrow {\mathbb R} \cup \{\infty\}$ is lower semicontinuous and bounded from below, then

\begin{equation*}\liminf_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF \circ S_n) \ge \sup_{x \in E}(F(x)-\Lambda^*(x)).\end{equation*}

Suppose also that the sub-level sets of $\alpha$ are pre-compact subsets of $\mathcal{P}_\psi(E)$ , for $\psi(x) \,:\!= \|x\|$ . If $F \colon E \rightarrow {\mathbb R} \cup \{-\infty\}$ is upper semicontinuous and bounded from above, then

\begin{equation*}\limsup_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF \circ S_n) \le \sup_{x \in E}(F(x)-\Lambda^*(x)).\end{equation*}

The proof makes use of a proposition, interesting in its own right, which generalizes the well-known result that the functions

\[t \mapsto \log\int_{\mathbb R} {\mathrm{e}}^{tx}\,\mu({\mathrm{d}} x) \quad \text{and} \quad t \mapsto \inf \bigg\{H(\nu \mid \mu) \colon \nu \in \mathcal{P}({\mathbb R}), \ \int_{\mathbb R} x\,\nu({\mathrm{d}} x) = t \bigg\}\]

are convex conjugates of each other (see e.g. [Reference Dupuis and Ellis22, Lemma 3.3.3]).

Proposition 2.1. Let $(E,\|\cdot\|)$ be a separable Banach space, and let $\psi(x) = \|x\|$ . Suppose the sub-level sets of $\alpha$ are pre-compact subsets of $\mathcal{P}_\psi(E)$ . Define $\Psi \colon E \rightarrow {\mathbb R} \cup \{\infty\}$ by

\[\Psi(x) = \inf\bigg\{\alpha(\nu) \colon \nu \in \mathcal{P}_\psi(E), \ \int_Ez\,\nu({\mathrm{d}} z) = x\bigg\},\]

where the integral is in the sense of Bochner. Define $\Psi^*$ on the continuous dual $E^*$ by

\[\Psi^*(x^*) = \sup_{x \in E}(\langle x^*,x\rangle - \Psi(x)).\]

Then $\Psi$ is convex and lower semicontinuous, and $\Psi^*(x^*) = \rho(x^*)$ for every $x^* \in E^*$ . In particular,

(2.12) \begin{equation}\Psi(x) = \sup_{x^* \in E^*}(\langle x^*,x\rangle - \rho(x^*)).\end{equation}

Proof. We first show that $\Psi$ is convex. Let $t \in (0,1)$ and $x_1,x_2 \in E$ . Fix $\epsilon \gt 0$ , and find $\nu_1,\nu_2 \in \mathcal{P}_\psi(E)$ such that $\int_Ez\nu_i({\mathrm{d}} z)=x_i$ and $\alpha(\nu_i) \le \Psi(x_i) + \epsilon$ . Convexity of $\alpha$ yields

\begin{align*} \Psi(tx_1 + (1-t)x_2) &\le \alpha(t\nu_1+(1-t)\nu_2) \\* & \le t\alpha(\nu_1) + (1-t)\alpha(\nu_2) \\* &\le t\Psi(x_1) + (1-t)\Psi(x_2) + \epsilon.\end{align*}

To prove that $\Psi$ is lower semicontinuous, first note that $\Psi$ is bounded from below since $\alpha$ is. Let $x_n \rightarrow x$ in E, and find $\nu_n \in \mathcal{P}_\psi(E)$ such that $\alpha(\nu_n) \le \Psi(x_n) + 1/n$ and $\int_Ez\nu_n({\mathrm{d}} z)=x_n$ for each n. Fix a subsequence $\{x_{n_k}\}$ such that $\Psi(x_{n_k}) \lt \infty$ for all k and $\Psi(x_{n_k})$ converges to a finite value (if no such subsequence exists, then there is nothing to prove, as $\Psi(x_n) \rightarrow \infty$ ). Then $\sup_{k}\alpha(\nu_{n_k}) \lt \infty$ , and because $\alpha$ has $\psi$ -weakly compact sub-level sets there exists a further subsequence (again denoted $n_k$ ) and some $\nu_\infty \in \mathcal{P}_\psi(E)$ such that $\nu_{n_k}\rightarrow\nu_\infty$ . The convergence $\nu_{n_k}\rightarrow\nu_\infty$ in the $\psi$ -weak topology implies

\[x = \lim_{k\rightarrow\infty}x_{n_k} = \lim_{k\rightarrow\infty}\int_Ez\nu_{n_k}({\mathrm{d}} z) = \int_Ez\,\nu_\infty({\mathrm{d}} z).\]

Using lower semicontinuity of $\alpha$ we conclude

(2.13) \begin{equation}\Psi(x) \le \alpha(\nu_\infty) \le \liminf_{k\rightarrow\infty}\alpha(\nu_{n_k}) \le \liminf_{k\rightarrow\infty}\Psi(x_{n_k}).\end{equation}

For every sequence $(x_n)$ in E and any subsequence thereof, this argument shows that there exists a further subsequence for which (2.13) holds, and this proves that $\Psi$ is lower semicontinuous. Next, compute $\Psi^*$ as follows:

\begin{align*}\Psi^*(x^*) &= \sup_{x \in E} (\langle x^*,x\rangle - \Psi(x)) \\* &= \sup_{x \in E}\,\sup\bigg\{\langle x^*,x\rangle - \alpha(\nu) \colon \nu \in \mathcal{P}_\psi(E), \ \int_Ez\nu({\mathrm{d}} z)=x \bigg\} \\ &= \sup_{\nu \in \mathcal{P}_\psi(E)}\bigg(\bigg\langle x^*,\int_Ez\nu({\mathrm{d}} z)\bigg\rangle - \alpha(\nu)\bigg) \\ &= \sup_{\nu \in \mathcal{P}_\psi(E)}\bigg(\int_E\langle x^*,z\rangle\nu({\mathrm{d}} z) - \alpha(\nu)\bigg) \\* &= \rho(x^*).\end{align*}

Indeed, we can take the supremum equivalently over $\mathcal{P}_\psi(E)$ or over $\mathcal{P}(E)$ in the last two steps, thanks to the assumption that $\alpha = \infty$ away from $\mathcal{P}_\psi(E)$ and our convention $\infty-\infty=-\infty$ . Because $\Psi$ is lower semicontinuous and convex, we conclude from the Fenchel–Moreau theorem [Reference Zalinescu55, Theorem 2.3.3] that it is equal to its biconjugate, which is precisely what (2.12) says.

Proof of Theorem 2.4. The map

\[\mathcal{P}_\psi(E) \ni \mu \mapsto F\bigg(\int_Ez\,\mu({\mathrm{d}} z)\bigg)\]

is upper (resp. lower) semicontinuous as soon as F is upper (resp. lower) semicontinuous. The claims then follow from Theorem 2.3 and Proposition 2.1.

3.1. Cramér’s condition

A first useful result is a condition under which the effective domain of $\alpha$ is contained in $\mathcal{P}_\psi(E)$ .

Proof. By definition, for each $\nu \in \mathcal{P}(E)$ ,

\[\infty \gt \rho(\lambda\psi) \ge \lambda\int \psi\,{\mathrm{d}} \nu - \alpha(\nu).\]

If $\alpha(\nu) \lt \infty$ then certainly $\int\psi\,{\mathrm{d}} \nu \lt \infty$ .

The next and more important proposition identifies a condition under which the sub-level sets of $\alpha$ are not only weakly compact (which we always assume) but also $\psi$ -weakly compact.

Proposition 3.2. Fix a continuous function $\psi \colon E \rightarrow {\mathbb R}_+$ . Suppose

(3.1) \begin{equation}\lim_{m\rightarrow\infty}\rho(\lambda\psi 1_{\{\psi \ge m\}}) = \rho(0) \quad \text{{for all} } \lambda \gt 0.\end{equation}

Then, for each $c \in {\mathbb R}$ , the weak and $\psi$ -weak topologies coincide on $\{\nu \in \mathcal{P}(E) \colon \alpha(\nu) \le c\} \subset \mathcal{P}_\psi(E)$ ; in particular, the sub-level sets of $\alpha$ are $\psi$ -weakly compact.

A first step in the proof comes from the following simple lemma, worth stating separately for emphasis.

Proof. The second claim is just Proposition 3.1. For $m,\lambda \gt 0$ we have $\lambda\psi \le \lambda m + \lambda\psi1_{\{\psi \ge m\}}$ , and thus properties (R1) and (R2) of Theorem 2.1 imply

\[\rho(\lambda \psi) \le \lambda m + \rho(\lambda\psi 1_{\{\psi \ge m\}}).\]

By (3.1), for m sufficiently large the right-hand side is finite.

Proof of Proposition 3.2. Fix $c \in {\mathbb R}$ , and abbreviate $S = \{\nu \in \mathcal{P}(E) \colon \alpha(\nu) \le c\}$ . Assume $S \neq \emptyset$ . Note that Lemma 3.1 implies $S \subset \mathcal{P}_\psi(E)$ . It suffices to prove that the map $\nu \mapsto \int_Ef\,{\mathrm{d}} \nu$ is weakly continuous on S for every continuous $f \colon E \to {\mathbb R}$ with $|\,f| \le 1 + \psi$ . Note that for $\eta_n,\eta \in \mathcal{P}({\mathbb R})$ with $\eta_n\to\eta$ weakly, we have $\int g\,{\mathrm{d}}\eta_n \to \int g\,{\mathrm{d}}\eta$ for each continuous function g which is uniformly integrable, in the sense that

\[\lim_{m\rightarrow\infty}\sup_n\int_{\{|g| \ge m\}}|g|\,{\mathrm{d}}\eta_n = 0.\]

(See [Reference Dupuis and Ellis22, Theorem A.3.19].) Applying this to the image measures $\{\nu \circ f^{-1} \colon \nu \in S\}$ for f as above, we find that it suffices to prove the uniform integrability condition

\[\lim_{m\rightarrow\infty}\sup_{\nu \in S}\int_{\{\psi \ge m\}}\psi\,{\mathrm{d}} \nu = 0.\]

By definition of $\rho$ , for $m \gt 0$ and $\nu \in S$ ,

(3.2) \begin{equation}\lambda\int_{\{\psi \ge m\}}\psi\,{\mathrm{d}} \nu \le \rho(\lambda\psi 1_{\{\psi \ge m\}}) + \alpha(\nu) \le \rho(\lambda\psi 1_{\{\psi \ge m\}}) + c {.}\end{equation}

Given $\epsilon \gt 0$ , choose $\lambda \gt 0$ large enough that $(\epsilon + \rho(0) + c)/\lambda \le \epsilon$ . Then choose m large enough that $\rho(\lambda\psi 1_{\{\psi \ge m\}}) \le \epsilon + \rho(0)$ , which is possible because of assumption (3.1). It then follows from (3.2) that $\int_{\{\psi \ge m\}}\psi\,{\mathrm{d}} \nu \le \epsilon$ , and the proof is complete.

We refer to (3.1) as the strong Cramér condition. Several extensions of the classical form of Sanov’s theorem to stronger topologies rely on what might be called a ‘strong Cramér condition’. For instance, if $\psi \colon E \rightarrow {\mathbb R}_+$ is continuous, the results of Schied [Reference Schied50] indicate that Sanov’s theorem can be extended to the $\psi$ -weak topology if (and essentially only if) $\log\int_E {\mathrm{e}}^{\lambda\psi}\,{\mathrm{d}} \mu \lt \infty$ for every $\lambda \ge 0$ ; see also [Reference Wang, Wang and Wu53] and [Reference Eichelsbacher and Schmock24].

The form of our strong Cramér condition (3.1) was heavily inspired by the work of Owari [Reference Owari44] on continuous extensions of monotone convex functionals. In several cases of interest (namely, Propositions 4.1 and 5.3 below), it turns out that a converse to Lemma 3.1 is true, that is, the strong Cramér condition (3.1) is equivalent to the statement that $\rho(\lambda\psi) \lt \infty$ for all $\lambda \gt 0$ . In general, however, the strong Cramér condition is the strictly stronger statement. Consider the following simple example, borrowed from [Reference Owari44, Example 3.7]. Let $E = \{0,1,\ldots,\}$ be the natural numbers, and define $\mu_n \in \mathcal{P}(E)$ by $\mu_1\{0\}=1$ , $\mu_n\{0\} = 1-1/n$ , and $\mu_n\{n\} = 1/n$ . Let M denote the closed convex hull of $(\mu_n)$ . Then M is convex and weakly compact. Define $\alpha(\mu) = 0$ for $\mu \in M$ and $\alpha(\mu)=\infty$ otherwise. Then $\alpha$ satisfies our standing assumptions, and $\rho(\,f) = \sup_{\mu \in M}\int f\,{\mathrm{d}} \mu = \sup_n\int f\,{\mathrm{d}} \mu_n$ . Finally, let $\psi(x)=x$ for $x \in E$ . Then $\rho(\lambda\psi) = \lambda \lt \infty$ because $\int \psi\,{\mathrm{d}} \mu_n = 1$ for all n, and similarly $\rho(\lambda \psi1_{\{\psi \ge m\}}) = \lambda$ because $\int\psi1_{\{\psi \ge m\}}\,{\mathrm{d}} \mu_n = 1_{\{n \ge m\}}$ . In particular, $\rho(\lambda\psi) \lt \infty$ for all $\lambda \gt 0$ , but the strong Cramér condition fails.

Finally, we remark that it is conceivable that a converse to Proposition 3.1 might hold, that is, the strong Cramér condition (3.1) may be equivalent to the pre-compactness of the sub-level sets of $\alpha$ in $\mathcal{P}_\psi(E)$ . Indeed, the results of Schied [Reference Schied50, Theorem Reference Agueh and Carlier2] and Owari [Reference Owari44, Theorem 3.8] suggest that this may be the case. But this remains an open problem.

3.2. Implications of $\psi$ -weakly compact sub-level sets

This section contains two results to be used occasionally below. First is a useful lemma that aids in the computation of $\rho(\,f)$ for certain unbounded f in Section 4.

Lemma 3.2. If $f \colon E \to {\mathbb R}$ is upper semicontinuous and bounded from above, then

\begin{equation*}\rho(\,f) = \lim_{m\to\infty}\rho(\,f \vee ({-}m)) = \inf_{m \ge 0}\rho(\,f \vee ({-}m)).\end{equation*}

If $f \colon E \to {\mathbb R}$ is measurable and bounded from below, then

\begin{equation*}\rho(\,f) = \lim_{m\to\infty}\rho(\,f \wedge m) = \sup_{m \ge 0}\rho(\,f \wedge m).\end{equation*}

Lastly, let $\psi \colon E \rightarrow {\mathbb R}_+$ be continuous. If the sub-level sets of $\alpha$ are pre-compact subsets of $\mathcal{P}_\psi(E)$ , and if $f \colon E \to {\mathbb R}$ is measurable with $f \ge -c(1+\psi)$ pointwise for some $c \ge 0$ , then

\begin{equation*}\rho(\,f) = \lim_{m\to\infty}\rho(\,f \wedge m) = \sup_{m \ge 0}\rho(\,f \wedge m).\end{equation*}

Proof. The second claim is a special case of the final claim with $\psi \equiv 0$ . To prove the final claim, note first that $\rho(\,f \wedge m)$ is non-decreasing in m (see (R1) of Theorem 2.1). We find

\begin{align*} \rho(\,f) &= \sup_{\nu \in \mathcal{P}_\psi(E)}\bigg(\int_E f\,{\mathrm{d}} \nu - \alpha(\nu)\bigg) = \sup_{m \ge 0}\sup_{\nu \in \mathcal{P}_\psi(E)}\bigg(\int_E f \wedge m\,{\mathrm{d}} \nu - \alpha(\nu)\bigg) = \sup_{m \ge 0}\rho(\,f \wedge m).\end{align*}

Indeed, for each $\nu \in \mathcal{P}_\psi(E)$ , the monotone convergence theorem applies because $f \wedge m$ for $m \ge 0$ are bounded from below by the $\nu$ -integrable function $-c(1+\psi)$ . To prove the first claim, abbreviate $f_m= f \vee ({-}m)$ for $m \ge 0$ . Monotonicity of $\rho$ implies $\inf_{m \ge 0}\rho(\,f_m) \ge \rho(\,f)$ , so we need only prove the reverse inequality. Assume without loss of generality that $\inf_{m \ge 0}\rho(\,f_m) \gt -\infty$ . For each n, we may find for each n some $\nu_n \in \mathcal{P}_\psi(E)$ such that

(3.3) \begin{equation}-\infty < \inf_{m \ge 0}\rho(\,f_m) \le \rho(\,f_n) \le \int_Ef_n\,{\mathrm{d}} \nu_n - \alpha(\nu_n) + 1/n.\end{equation}

This implies $\sup_n\alpha(\nu_n) \lt \infty$ , because f is bounded from above. Pre-compactness of the sub-level sets of $\alpha$ allows us to extract a subsequence ${n_k}$ and $\nu_\infty \in \mathcal{P}(E)$ such that $\nu_{n_k} \rightarrow \nu_\infty$ weakly. By Skorokhod’s representation, we may construct random variables $X_k$ and $X_\infty$ with respective laws $\nu_{n_k}$ and $\nu_\infty$ such that $X_k \rightarrow X_\infty$ a.s. The upper semicontinuity assumption implies $\limsup_{k\rightarrow\infty}f_{n_k}(X_k) \le f(X_\infty)$ almost surely. We then conclude from Fatou’s lemma that

\[\limsup_{k\rightarrow\infty}\int_Ef_{n_k}\,{\mathrm{d}} \nu_{n_k} = \limsup_{k\rightarrow\infty}{\mathbb E}[\,f_{n_k}(X_k)] \le {\mathbb E}[\,f(X_\infty)] = \int_Ef\,{\mathrm{d}} \nu_\infty.\]

Since $\alpha$ is weakly lower semicontinuous, we conclude from (3.3) that

\begin{equation}\inf_{m \ge 0}\rho(\,f_m) \le \int_Ef\,{\mathrm{d}} \nu_\infty - \alpha(\nu_\infty) \le \sup_{\nu \in \mathcal{P}(E)}\bigg(\int_Ef\,{\mathrm{d}} \nu - \alpha(\nu)\bigg) = \rho(\,f).\end{equation}

4.1. Shortfall risk measures

Fix $\mu \in \mathcal{P}(E)$ and a non-decreasing, non-constant, convex function $\ell \colon {\mathbb R} \rightarrow {\mathbb R}_+$ satisfying $\ell(x) \lt 1$ for all $x \lt 0$ . Let $\ell^*(y) = \sup_{x \in {\mathbb R}}(xy - \ell(x))$ denote the convex conjugate, and define $\alpha \colon \mathcal{P}(E) \rightarrow [0,\infty]$ by

(4.1) \begin{align}\alpha(\nu) = \begin{cases}\displaystyle \inf_{t \gt 0}\dfrac{1}{t}\bigg(1 + \int_E\ell^*\bigg(t\dfrac{{\mathrm{d}} \nu}{{\mathrm{d}} \mu}\bigg)\,{\mathrm{d}} \mu\bigg) &\text{if } \nu \ll \mu, \\[7pt]\infty &\text{otherwise}.\end{cases}\end{align}

Note that $\ell^*(x) \ge - \ell(0) \ge -1$ , by assumption and by continuity of $\ell$ , so that $\alpha \ge 0$ . Define $\rho$ as usual by (1.1). It is known [Reference Föllmer and Schied28, Proposition 4.115] that, for $f \in B(E)$ ,

(4.2) \begin{equation}\rho(\,f) = \inf\bigg\{m \in {\mathbb R} \colon \int_E\ell(\,f(x)-m)\mu({\mathrm{d}} x) \le 1\bigg\}.\end{equation}

Refer to the book of Föllmer and Schied [Reference Föllmer and Schied28, Section 4.9] for a thorough study of the properties of $\rho$ . Notably, they show that $\rho$ satisfies all of properties (R1–R4) of Theorem 2.1, and that both dual formulas hold:

\begin{equation*}\rho(\,f) = \sup_{\nu \in \mathcal{P}(E)}\bigg(\int_Ef\,{\mathrm{d}} \nu - \alpha(\nu)\bigg), \quad\alpha(\nu) = \sup_{f \in B(E)}\bigg(\int_Ef\,{\mathrm{d}} \nu - \rho(\,f)\bigg).\end{equation*}

If $\ell(x)= {\mathrm{e}}^x$ we recover $\rho(\,f)=\log\int_E {\mathrm{e}}^{\,f}\,{\mathrm{d}} \mu$ and $\alpha(\nu) = H(\nu \mid \mu)$ . If $\ell(x) = [(1+x)^+]^q$ for some $q \ge 1$ , then

(4.3) \begin{equation}\alpha(\nu) = \|{\mathrm{d}} \nu/{\mathrm{d}} \mu\|_{L^p(\mu)}-1\quad \text{for } \nu \ll \mu, \qquad \alpha(\nu) = \infty \quad \text{otherwise},\end{equation}

where $p=q/(q-1)$ , and where of course

\[\|\,f\|_{L^p(\mu)} = \bigg(\int |\,f|^p\,{\mathrm{d}} \mu\bigg)^{1/p}{.}\]

See [Reference Föllmer and Schied28, Example 4.118] or [Reference Lacker39, Section 3.1] for this computation. The $-1$ is a convenient normalization, ensuring that $\alpha(\nu)=0$ if and only if $\nu=\mu$ .

In the rest of this subsection we work with $\alpha$ and $\rho$ given as in (4.1) and (4.2). The following result shows how the strong Cramér condition (3.1) simplifies in the present context. It is essentially contained in [Reference Owari44, Proposition 7.3], but we include the short proof.

Proposition 4.1. Let $\psi \colon E \rightarrow {\mathbb R}_+$ be measurable. Suppose

\[ \int_E\ell(\lambda\psi(x))\mu({\mathrm{d}} x) < \infty\quad {for\ all }\ \lambda \gt 0.\]

Then the strong Cramér condition holds,

\[ \lim_{m\rightarrow\infty}\rho(\lambda\psi 1{\{\psi \ge m\}})\rightarrow 0\quad {for\ each }\ \lambda \gt 0.\]

In particular, the sub-level sets of $\alpha$ are compact subsets of $\mathcal{P}_\psi(E)$ .

Proof. The final claim is simply an application of Proposition 3.2. Fix $\epsilon \gt 0$ and $\lambda \gt 0$ . Since $\ell$ is non-decreasing, the following two limits hold:

\begin{equation*}\lim_{m\rightarrow\infty}\mu(\psi < m) = 1, \quad \lim_{m\rightarrow\infty}\int_{\{\psi \ge m\}}\ell (\lambda\psi(x) - \epsilon )\mu({\mathrm{d}} x) = 0.\end{equation*}

Since $\ell({-}\epsilon) \lt 1$ , it follows that, for sufficiently large m,

\begin{align*}1 &\ge \ell({-}\epsilon)\mu(\psi < m) + \int_{\{\psi \ge m\}}\ell (\lambda\psi(x) - \epsilon )\mu({\mathrm{d}} x) \\* &= \int_E\ell(\lambda\psi(x)1_{\{\psi \ge m\}}(x) - \epsilon)\mu({\mathrm{d}} x).\end{align*}

Next, the second assertion of Lemma 3.2 implies $\rho(\lambda\psi 1_{\{\psi \ge m\}}) = \sup_{n\ge 0}\rho(n \wedge (\lambda\psi 1_{\{\psi \ge m\}}))$ . For each n, we use the identity (4.2), which is valid for bounded f, to get, for sufficiently large m,

\begin{align}\rho(\lambda\psi 1_{\{\psi \ge m\}}) &= \sup_{n\ge 0}\rho(n \wedge (\lambda\psi 1_{\{\psi \ge m\}})) \nonumber\\ &= \sup_{n \ge 0}\inf\bigg\{c \in {\mathbb R} \colon \int_E\ell (n \wedge (\lambda\psi 1_{\{\psi \ge m\}}) - c )\,{\mathrm{d}} \mu \le 1 \bigg\} \nonumber\displaybreak\\ &\le \inf\bigg\{c \in {\mathbb R} \colon \int_E\ell ( \lambda\psi 1_{\{\psi \ge m\}} - c )\,{\mathrm{d}} \mu \le 1 \bigg\} \nonumber\\ &\le \epsilon.\end{align}

Note that (4.2) is only valid, a priori, for bounded f, although the expression on the right-hand side certainly makes sense for unbounded f. The next results provide some useful cases for which the identity (4.2) carries over to unbounded functions, and these will be needed in the proof of Corollary 1.2. In the following, define $\ell(\pm \infty) = \lim_{x \rightarrow \pm\infty}\ell(x)$ .

Proof. Let H(f) denote the right-hand side of (4.2), well-defined for any measurable function $f \colon E \to {\mathbb R}$ . We must show $\rho(\,f)=H(\,f)$ for f as in the statement of the proposition. As was mentioned above, it is known from [Reference Föllmer and Schied28, Proposition 4.115] that $\rho(\,f)=H(\,f)$ for bounded f.

Step 1. Assume first that f is continuous and bounded from above, with $|\,f| \le c(1+\psi)$ . Let $f_n = f \vee ({-}n)$ for $n \ge 0$ . Since $f_n$ is bounded for each n, we have $\rho(\,f_n)=H(\,f_n)$ . The first assertion of Lemma 3.2 then implies $\rho(\,f) = \lim_{n\to\infty}\rho(\,f_n) = \lim_{n\to\infty}H(\,f_n)$ . It remains to show $H(\,f_n)\to H(\,f)$ . Clearly $H(\,f_n) \ge H(\,f_{n+1}) \ge H(\,f)$ for each n since $f_n \ge f_{n+1}$ pointwise and $\ell$ is non-decreasing, so the sequence $H(\,f_n)$ has a limit. As $\ell$ is continuous and strictly increasing in a neighborhood of the origin, note that H(f) is the unique solution $c \in {\mathbb R}$ of the equation

(4.4) \begin{equation}\int_E\ell(\,f(x)-c)\mu({\mathrm{d}} x) = 1.\end{equation}

Similarly, $H(\,f_n)$ uniquely solves

\[\int_E\ell(\,f_n(x)-H(\,f_n))\mu({\mathrm{d}} x) = 1.\]

Let $c = \lim_{n\to\infty}H(\,f_n)$ , and note that the integrands $\ell(\,f_n(x)-H(\,f_n))$ are uniformly bounded and converge pointwise to $\ell(\,f(x)-c)$ . Passing to the limit using dominated convergence shows that c solves the equation (4.4), which implies $c=H(\,f)$ .

Step 2. We now turn to general continuous f satisfying $|\,f| \le c(1+\psi)$ . Define $f_n = f \wedge n$ for $n \ge 0$ , so that $f_n$ is bounded from above. By Step 2, $\rho(\,f_n)=H(\,f_n)$ for each n. By Proposition 4.1, the sub-level sets of $\alpha$ are $\psi$ -weakly compact, and the third assertion of Lemma 3.2 yields $\rho(\,f_n) \to \rho(\,f)$ . It remains to show $H(\,f_n)\to H(\,f)$ . To see this, note first that $H(\,f_n) \le H(\,f_{n+1}) \le H(\,f)$ for each n since $f_n \le f_{n+1}$ pointwise. Let $\epsilon \gt 0$ and $c = H(\,f)-\epsilon$ , and note that the definition of H and monotonicity of $\ell$ imply $\int_E\ell(\,f(x) - c)\mu({\mathrm{d}} x) \gt 1$ . By monotone convergence, there exists n such that $\int_E\ell(\,f_n(x) - c)\mu({\mathrm{d}} x) \gt 1$ . The definition of H now implies $H(\,f_n) \gt c = H(\,f) - \epsilon$ . As $\epsilon \gt 0$ was arbitrary, we conclude $H(\,f_n)\to H(\,f)$ .

We record here for later use a simple but useful lemma.

4.2. Proofs of Theorem 1.2, Corollary 1.1, and Corollary 1.2

With these generalities in hand, we now turn toward the proof of Theorem 1.2. The idea is to apply Theorem 2.3 with $\alpha$ defined as in (4.3). The following estimate is crucial.

Lemma 4.2. Let $q \in (1,\infty]$ , and let $p=q/(q-1)$ denote the conjugate exponent. Let $\alpha$ be as in (4.3). Then, for each $n \ge 1$ and $\nu \in \mathcal{P}(E^n)$ with $\nu \ll \mu^n$ ,

(4.5) \begin{equation}\alpha_n(\nu) \le n^{1/q}\|{\mathrm{d}} \nu/{\mathrm{d}} \mu^n\|_{L^p(\mu^n)}.\end{equation}

Proof. The case $p=\infty$ and $q=1$ follows by sending $p \rightarrow\infty$ in (4.5), so we prove only the case $p \lt \infty$ . As we will be working with conditional expectations, it is convenient to work with a more probabilistic notation. Fix n, and endow $\Omega = E^n$ with its Borel $\sigma$ -field as well as the probability $P = \mu^n$ . Let $X_i \colon E^n \rightarrow E$ denote the natural projections, and let $\mathcal{F}_k = \sigma(X_1,\ldots,X_k)$ denote the natural filtration, for $k=1,\ldots,n$ , with $\mathcal{F}_0 \,:\!= \{\emptyset,\Omega\}$ . For $\nu \in \mathcal{P}(E^n)$ and $k=1,\ldots,n$ , let $\nu_k$ denote a version of the regular conditional law of $X_k$ given $\mathcal{F}_{k-1}$ under $\nu$ , or symbolically $\nu_k \,:\!= \nu(X_k \in \cdot \mid \mathcal{F}_{k-1})$ . Let ${\mathbb E}^\nu$ denote integration with respect to $\nu$ . Since $P(X_k \in \cdot \mid \mathcal{F}_{k-1}) = \mu$ a.s., if $\nu \ll P$ then

\[\dfrac{{\mathrm{d}} \nu_k}{{\mathrm{d}} \mu} = \dfrac{{\mathbb E}^P[{\mathrm{d}} \nu/{\mathrm{d}} P \mid \mathcal{F}_k]}{{\mathbb E}^P[{\mathrm{d}} \nu/{\mathrm{d}} P \mid \mathcal{F}_{k-1}]} =\!:\, \dfrac{M_k}{M_{k-1}} \ \ \text{a.s.}, \quad\text{where } \dfrac{0}{0} \,:\!= 0.\]

Therefore

\[\alpha(\nu_k) = {\mathbb E}^P\bigg[\bigg(\dfrac{M_k}{M_{k-1}}\bigg)^p\mid \mathcal{F}_{k-1}\bigg]^{1/p}-1.\]

Note that $(M_k)_{k=0}^n$ is a non-negative martingale, with $M_0 = 1$ and $M_n = {\mathrm{d}} \nu/{\mathrm{d}} P$ . Then

\begin{align*} \alpha_n(\nu) &= {\mathbb E}^\nu\Bigg[\sum_{k=1}^n\alpha(\nu_k)\Bigg] \\* &= {\mathbb E}^P\Bigg[M_n\sum_{k=1}^n\bigg({\mathbb E}^P\bigg[\bigg(\dfrac{M_k}{M_{k-1}}\bigg)^p\mid \mathcal{F}_{k-1}\bigg]^{1/p}-1\bigg)\Bigg] \\* &= {\mathbb E}^P\Bigg[\sum_{k=1}^n ({\mathbb E}^P[ M_k^p \mid \mathcal{F}_{k-1}]^{1/p}-M_{k-1} )\Bigg].\end{align*}

Subadditivity of $x \mapsto x^{1/p}$ implies

\[({\mathbb E}^P[M_k^p \mid \mathcal{F}_{k-1}])^{1/p} \le ({\mathbb E}^P[M_k^p - M_{k-1}^p \mid \mathcal{F}_{k-1}])^{1/p} + M_{k-1},\]

where the right-hand side is well-defined because

\[{\mathbb E}^P[M_k^p \mid \mathcal{F}_{k-1}] \ge{\mathbb E}^P[M_k \mid \mathcal{F}_{k-1}]^p = M_{k-1}^p.\]

Concavity of $x \mapsto x^{1/p}$ and Jensen’s inequality yield

\begin{align} \alpha_n(\nu) & \le {\mathbb E}^P\Bigg[\sum_{k=1}^n ({\mathbb E}^P[M_k^p - M_{k-1}^p \mid \mathcal{F}_{k-1}])^{1/p}\Bigg] \nonumber\\ &\le n^{1-{1}/{p}}\Bigg({\mathbb E}^P\Bigg[\sum_{k=1}^n{\mathbb E}^P[M_k^p - M_{k-1}^p \mid \mathcal{F}_{k-1}]\Bigg]\Bigg)^{1/p} \nonumber\\ &= n^{1/q}({\mathbb E}^P[M_n^p - M_0^p])^{1/p} \nonumber\\ &\le n^{1/q}({\mathbb E}^P[M_n^p])^{1/p}.\end{align}

Proof of Theorem 1.2. Again, let $q \in (1,\infty)$ and $p=q/(q-1)$ , and let $\alpha$ be as in (4.3), noting that it corresponds to (4.1) with $\ell(x) = [(1+x)^+]^q$ . Then Proposition 4.1 and the assumption that $\int\psi^q\,{\mathrm{d}} \mu \lt \infty$ imply that the sub-level sets of $\alpha$ are pre-compact subsets of $\mathcal{P}_\psi(E)$ . Hence, Theorem 2.3 applies to the $\psi$ -weakly upper semicontinuous function $F \colon \mathcal{P}_\psi(E) \rightarrow [-\infty,0]$ defined by $F(\nu) = 0$ if $\nu \in A$ and $F(\nu) = -\infty$ otherwise. This yields

(4.6) \begin{equation}\limsup_{n\rightarrow\infty}\dfrac{1}{n}\rho_n(nF \circ L_n) \le -\inf_{\nu \in A}\alpha(\nu).\end{equation}

Now use Lemma 4.2, noting that $({1}/{n})n^{1/q} = n^{-1/p}$ , to get

\begin{align*}\dfrac{1}{n}\rho_n(nF\circ L_n) &= \sup_{\nu \in \mathcal{P}(E^n)}\bigg(\int_{E^n} F\circ L_n\,{\mathrm{d}} \nu- \dfrac{1}{n}\alpha_n(\nu)\bigg) \\* &= -\inf\bigg\{\dfrac{1}{n}\alpha_n(\nu) \colon \nu \in \mathcal{P}(E^n), \ \nu(L_n \in A)=1\bigg\} \\* &\ge -\inf\{n^{-1/p}\|{\mathrm{d}} \nu/{\mathrm{d}} \mu^n\|_{L^p(\mu^n)} \colon \nu \in \mathcal{P}(E^n), \ \nu \ll \mu^n, \ \nu(L_n \in A)=1\}.\end{align*}

Set $B_n = \{x \in E^n \colon L_n(x) \in A\}$ , and define $\nu \ll \mu^n$ by ${\mathrm{d}} \nu/{\mathrm{d}} \mu^n = 1_{B_n}/\mu^n(B_n)$ . A quick computation yields

\[\|{\mathrm{d}} \nu/{\mathrm{d}} \mu^n\|_{L^p(\mu^n)} = \mu^n(B_n)^{(1-p)/p} = \mu^n(B_n)^{-1/q}.\]

Thus

\[\dfrac{1}{n}\rho_n(nF \circ L_n) \ge - (n^{1/p}\mu^n(B_n)^{1/q})^{-1}.\]

Combine this with (4.6) to get

\begin{equation*}\limsup_{n\rightarrow\infty}-(n^{1/p}\mu^n(L_n \in A)^{1/q})^{-1} \le -\inf_{\nu \in A}\alpha(\nu).\end{equation*}

Recalling the definition of $\alpha$ from (4.3) and noting that $q/p = q-1$ , this inequality can be rewritten as the desired result.

Proof of Corollary 1.1. Define a continuous function $\psi \colon E \to {\mathbb R}_+$ by $\psi(x) = d^r(x,x_0)$ . Note that $\mathcal{W}_r$ then metrizes $\mathcal{P}_\psi(E)$ (see [Reference Villani52, Theorem 7.12]). Hence, the set

\[A = \{\nu \in \mathcal{P}_\psi(E) \colon \mathcal{W}_r(\nu,\mu) \ge a\}\]

is closed in $\mathcal{P}_\psi(E)$ . Because $\int \psi^{q/r}\,{\mathrm{d}} \mu = \int d^q(x,x_0)\,\mu({\mathrm{d}} x) \lt \infty$ by assumption, we may apply Theorem 1.2 with $q/r$ in place of q to get

\begin{equation*}\limsup_{n\to\infty} n^{{q}/{r}-1}\mu^n(\mathcal{W}_r(L_n,\mu) \ge a) \le \bigg(\inf_{\nu \in A} \alpha(\nu)\bigg)^{-q/r},\end{equation*}

where $\alpha$ is defined as in (4.3) with $p=(q/r)/(q/r-1)$ . It remains to show that $\inf_{\nu \in A} \alpha(\nu) \gt 0$ . But this follows from Lemma 4.1, since A is closed, $\mu \notin A$ , and $\int \psi^{q/r}\,{\mathrm{d}} \mu \lt \infty$ .

Proof of Corollary 1.2. Again, let $\alpha$ be as in (4.3), and note that it corresponds to the shortfall risk measure (4.2) with $\ell(x) = [(1+x)^+]^q$ . Let $\psi(x) = \|x\|$ , and consider the $\mathcal{P}_\psi(E)$ -closed set

\[B = \bigg\{\mu \in \mathcal{P}_\psi(E) \colon \int_Ez\,\mu({\mathrm{d}} z) \in A\bigg\},\]

where the integral is defined in the Bochner sense. Proposition 4.1 and the assumption that $\int\psi^q\,{\mathrm{d}} \mu = {\mathbb E}[\|X_1\|^q] \lt \infty$ imply that the sub-level sets of $\alpha$ are pre-compact subsets of $\mathcal{P}_\psi(E)$ . We may then apply Theorem 1.2 to get

\[\limsup_{n\rightarrow\infty}n^{q-1}{\mathbb P}\Bigg(\dfrac{1}{n}\sum_{i=1}^nX_i \in A\Bigg) \le \bigg(\inf_{\nu \in B}\alpha(\nu)\bigg)^{-q},\]

where again $\alpha$ is as in (4.3). It remains to simplify the right-hand side. Proposition 2.1 yields

\[\sup_{x^* \in E^*}(\langle x^*,x\rangle - \rho(x^*)) = \inf\bigg\{\alpha(\nu) \colon \nu \in \mathcal{P}_\psi(E), \ \int_Ez\,\nu({\mathrm{d}} z)=x\bigg\}\quad \text{for } x \in E.\]

Infimize over $x \in A$ on both sides to get

(4.7) \begin{equation}\inf_{\nu \in B}\alpha(\nu) = \inf_{x \in A}\sup_{x^* \in E^*}(\langle x^*,x\rangle - \rho(x^*)).\end{equation}

According to Proposition 4.2, for $x^* \in E^*$ we have

\begin{equation*}\rho(x^*) = \inf\bigg\{m \in {\mathbb R} \colon \int_E [(1+x^*(x)-m)^+]^q\mu({\mathrm{d}} x) \le 1\bigg\} = \Lambda(x^*),\end{equation*}

where the latter equality is simply the definition of $\Lambda$ given in the statement of Corollary 1.2. Hence, the identity (4.7) becomes $\inf_{\nu \in B}\alpha(\nu) = \inf_{x \in A}\Lambda^*(x)$ , and the proof is complete.

4.3. Stochastic optimization with heavy tails

This section elaborates on the application discussed in Section 1.1.3, concerning the convergence of Monte Carlo estimates for stochastic optimization problems. We use the notation of Section 1.1.3, and we begin with the proof of Theorem 1.3.

Proof of Theorem 1.3. Let $A = \{\nu \in \mathcal{P}_\psi(E) \colon |V(\nu)-V(\mu)| \ge \epsilon\}$ . The map

\[\mathcal{X} \times \mathcal{P}_\psi(E) \ni (x,\nu) \mapsto \int_Eh(x,w)\nu({\mathrm{d}} x)\]

is jointly continuous. By Berge’s theorem [Reference Aliprantis and Border3, Theorem 17.31], V is continuous on $\mathcal{P}_\psi(E)$ , so A is closed. Theorem 1.2 implies

\begin{equation*}\limsup_{n\rightarrow\infty}n^{q-1}\mu^n(|V(L_n)-V(\mu)| \ge \epsilon) = \limsup_{n\rightarrow\infty}n^{q-1}\mu^n(L_n \in A) \le \bigg(\inf_{\nu \in A}\alpha(\nu)\bigg)^{-q}.\end{equation*}

Note that $q/p=q-1$ , and finally use Lemma 4.1 to conclude $\inf_{\nu \in A}\alpha(\nu) \gt 0$ .

Now we have shown that the optimal value itself converges, we turn to the convergence of optimizers themselves.

Theorem 4.1. Grant the assumptions of Theorem 1.3. Let $\hat{x} \colon \mathcal{P}_\psi(E) \rightarrow \mathcal{X}$ be any measurable function satisfying

\[\hat{x}(\nu) \in \arg\min_{x \in \mathcal{X}}\int_Eh(x,w)\nu({\mathrm{d}} w)\quad {for\ each\ }\ \nu.\]

Suppose there exist a measurable function $\varphi \colon {\mathbb R} \rightarrow {\mathbb R}$ and a compatible metric d on $\mathcal{X}$ such that

\[\varphi(d(\hat{x}(\mu),x)) \le \int_Eh(x,w)\mu({\mathrm{d}} w) - \int_Eh(\hat{x}(\mu),w)\mu({\mathrm{d}} w).\]

Then, for any $\epsilon \gt 0$ ,

\[\limsup_{n\rightarrow\infty}n^{q-1}\mu^n(\varphi(d(\hat{x}(\mu),\hat{x}(L_n))) \ge \epsilon) < \infty.\]

In particular, if $\varphi$ is strictly increasing with $\varphi(0)=0$ , then for any $\epsilon \gt 0$ ,

\[\limsup_{n\rightarrow\infty}n^{q-1}\mu^n(d(\hat{x}(\mu),\hat{x}(L_n)) \ge \epsilon) < \infty.\]

Such a function $\hat{x}$ exists because $(x,\nu) \mapsto \int_Eh(x,w)\nu({\mathrm{d}} w)$ is measurable in $\nu$ and continuous in x; see e.g. [Reference Aliprantis and Border3, Theorem 18.19].

Proof. Note that for $\epsilon \gt 0$ , on the event $\{\varphi(d(\hat{x}(\mu),\hat{x}(L_n))) \ge \epsilon\}$ we have

\begin{align*} \epsilon &\le \varphi(d(\hat{x}(\mu),\hat{x}(L_n))) \\* & \le \int_Eh(\hat{x}(L_n),w)\mu({\mathrm{d}} w) - \int_Eh(\hat{x}(\mu),w)\mu({\mathrm{d}} w) \\* & \le |V(L_n)-V(\mu)| + \sup_{x \in \mathcal{X}}\int_Eh(x,w)[\mu-L_n]({\mathrm{d}} w).\end{align*}

The first term converges at the right rate, thanks to Theorem 1.3, and it remains to check that

\[\limsup_{n\rightarrow\infty}n^{q-1}\mu^n\Bigg(\sup_{x \in \mathcal{X}}\int_Eh(x,w)[\mu-L_n]({\mathrm{d}} w) \ge \epsilon\Bigg) < \infty.\]

The map $(x,\nu) \mapsto \int_Eh(x,w)\nu({\mathrm{d}} w)$ is continuous on $\mathcal{X} \times \mathcal{P}_\psi(E)$ , so the map

\[\mathcal{P}_\psi(E) \ni \nu \mapsto \sup_{x \in \mathcal{X}}\int_Eh(x,w)[\mu-\nu]({\mathrm{d}} w)\]

is continuous by Berge’s theorem [Reference Aliprantis and Border3, Theorem 17.31]. Hence, the set

\[B \,:\!= \bigg\{\nu \in \mathcal{P}_\psi(E) \colon \sup_{x \in \mathcal{X}}\int_Eh(x,w)[\mu-\nu]({\mathrm{d}} w) \ge \epsilon\bigg\}\]

is closed in $\mathcal{P}_\psi(E)$ . Theorem 1.2 then implies

\begin{equation*}\limsup_{n\rightarrow\infty}n^{q-1}\mu^n\Bigg(\sup_{x \in \mathcal{X}}\int_Eh(x,w)[\mu-L_n]({\mathrm{d}} w) \ge \epsilon\Bigg) \le \bigg(\inf_{\nu \in B}\alpha(\nu)\bigg)^{-q},\end{equation*}

where $\alpha$ is defined as in (4.3). Finally, Lemma 4.1 implies that $\inf_{\nu \in B}\alpha(\nu) \gt 0$ .

Under the assumption $\int_E\psi^q\,{\mathrm{d}} \mu \lt \infty$ , we see that the value $V(L_n)$ always converges to $V(\mu)$ with the polynomial rate $n^{1-q}$ . To see when Theorem 4.1 applies, notice that in many situations, $\mathcal{X}$ is a convex subset of a normed vector space, and we have uniform convexity in the following form. There exists a strictly increasing function $\varphi$ such that $\varphi(0)=0$ and, for all $t \in (0,1)$ and $x,y \in \mathcal{X}$ ,

\begin{align*}&\int_Eh(tx + (1-t)y,w)\mu({\mathrm{d}} w) \\* & \quad\, \le t\int_Eh(x,w)\mu({\mathrm{d}} w) + (1-t)\int_Eh(y,w)\mu({\mathrm{d}} w) - t(1-t)\varphi(\|x-y\|).\end{align*}

See [Reference Kaniovski, King and Wets37, pages 202–203] for more on this.