2. Nuclearity of Bochner representable operators on $B(\Sigma)$

We will need the following result (see [Reference Khurana16, Theorem 3], [Reference Nowak20, Proposition 13 and Corollary 14]).

Grothendieck carried over the concept of nuclear operators to locally convex spaces [Reference Grothendieck12], [Reference Grothendieck13] (see also [Reference Yosida28, p. 289–293], [Reference Jarchow15, pp. 376–378], [Reference Schaefer24, Chap. 3, § 7], [Reference Trèves27, § 47]). Following [Reference Schaefer24, Chap. 3, § 7], [Reference Trèves27, § 47] and using Proposition 2.1 we have the following definition.

Definition 2.1. A linear operator $T:B(\Sigma)\rightarrow E$ between the locally convex space $(B(\Sigma),\tau(B(\Sigma),ca(\Sigma)))$ and a Banach space E is said to be nuclear if there exist a bounded and uniformly countably additive sequence $(\lambda_n)$ in $ca(\Sigma)$, a bounded sequence $(e_n)$ in E and a sequence $(\alpha_n)\in\ell^1$ such that

(2.1)\begin{equation} T(u)=\sum^\infty_{n=1}\alpha_n\left(\int_\Omega u(\omega)\, d\lambda_n\right)e_n \ \ \rm{for\, all}\ \ u\in B(\Sigma). \end{equation}

Then $T:B(\Sigma)\rightarrow E$ is $(\tau(B(\Sigma),ca(\Sigma)),\|\cdot\|_E)$-compact, that is, T(V) is relatively norm compact in E for some $\tau(B(\Sigma),ca(\Sigma))$-neighbourhood V of 0 in $B(\Sigma)$ (see [Reference Schaefer24, Chap. 3, § 7, Corollary 1], [Reference Trèves27, Theorem 47.3]). Hence T is $(\tau(B(\Sigma),ca(\Sigma)),$ $\|\cdot\|_E)$-continuous.

Let us put

\begin{equation*} \|T\|_{\tau-nuc}:=\inf\left\{\sum^\infty_{n=1} |\alpha_n|\,|\lambda_n|\,(\Omega)\,\|e_n\|_E\right\}, \end{equation*}

where the infimum is taken over all sequences $(\lambda_n)$ in $ca(\Sigma)$ and $(e_n)$ in E and $(\alpha_n)\in\ell^1$ such that T admits a representation (2.1).

According to [Reference Nowak19, Theorem 2.1] and Proposition 1.1 we have the following characterization of nuclear σ-smooth operators $T:B(\Sigma)\rightarrow E$.

Making us of [Reference Dinculeanu8, Sect.2, F, Theorem 30, p. 26] we have the following result.

Lemma 2.3. For $\mu\in ca(\Sigma)^+$ and $f\in L^1(\mu,E)$, let us put

\begin{equation*} \lambda(A):=\int_A\|f(\omega)\|_E\,d\mu,\quad\mbox{for all}\ \ A\in\Sigma, \end{equation*}

and

\begin{equation*} h_f(\omega):=f(\omega)/\|f(\omega)\|_E\ \ \mbox{if}\ \ f(\omega)\neq0\ \ \mbox{and}\ \ h_f(\omega):=0\ \ \mbox{if}\ \ f(\omega)=0. \end{equation*}

Then $h_f\in L^1(\lambda, E)$ and

\begin{equation*} \int_{\Omega}u(\omega)h_f(\omega)\,d\lambda=\int_{\Omega}u(\omega)f(\omega)\,d\mu,\quad \mbox{for all}\ \ u\in B(\Sigma). \end{equation*}

In particular, $\int_Ah_f(\omega)\,d\lambda=\int_Af(\omega)\,d\mu$ for all $A\in\Sigma$.

Proof. There exists a measure $\mu\in ca(\Sigma)^+$ and a function $f\in L^1(\mu,E)$ so that

\begin{equation*} T(u)=\int_\Omega u(\omega)f(\omega)\,d\mu,\quad \mbox{for all}\ \ u\in B(\Sigma). \end{equation*}

Hence

\begin{equation*} m_T(A)=\int_A f(\omega)\,d\mu\ \ \mbox{and}\ \ |m_T|(A)=\int_A\|f(\omega)\|_E\,d\mu,\quad \mbox{for all}\ \ A\in\Sigma, \end{equation*}

where m_T is a countably additive measure (see [Reference Diestel and Uhl7, Theorem 4, p. 46]), and in view of Proposition 1.1 T is σ-smooth. Hence using Lemma 2.3 we get

\begin{equation*} m_T(A)=\int_Af(\omega)\,d\mu=\int_Ah_f(\omega)\,d|m_T|,\quad \mbox{for all}\ \ A\in\Sigma, \end{equation*}

where $h_f\in L^1(|m_T|,E)$. By Theorem 2.2 we derive that T is a nuclear operator between the locally convex space $(B(\Sigma),\tau(B(\Sigma),ca(\Sigma)))$ and the Banach space E.

In view of Theorem 2.4 and Theorem 2.2 we can obtain the following characterization of Bochner representable operators $T:B(\Sigma)\rightarrow E$.

As a consequence of Theorem 2.4 and Theorem 2.2, we get

Corollary 2.6. Assume that $T:B(\Sigma)\rightarrow E$ is a Bochner representable operator. Then the mapping

\begin{equation*} T^*:E'\ni e'\mapsto e'\circ m_T\in ca(\Sigma) \end{equation*}

is a nuclear operator and $\|T^*\|_{nuc}=\|T\|_{nuc}=|m_T|(\Omega)$.

Proof. Let ɛ > 0 be given. In view of Theorem 2.4 and Theorem 2.2 there exist a bounded and uniformly countably additive sequence $(\lambda_n)$ in $ca(\Sigma)$, a bounded sequence $(e_n)$ in E and $(\alpha_n)\in\ell^1$ so that

\begin{equation*} T(u)=\sum^\infty_{n=1}\alpha_n \Phi_{\lambda_n}(u)\,e_n,\quad \mbox{for all} \ \ u\in B(\Sigma) \end{equation*}

and

(2.5)\begin{equation} \sum^\infty_{n=1}|\alpha_n|\;|\lambda_n|(\Omega)\;\|e_n\|_E\leq |m_T|(\Omega)+\varepsilon. \end{equation}

One can show that for each $e'\in E'$, we have

\begin{equation*} e'\circ T=\sum^\infty_{n=1}\alpha_n\,e'(e_n)\Phi_{\lambda_n} \ \ \mbox{in}\ \ B(\Sigma)'. \end{equation*}

Moreover, for each $e'\in E'$, we have $e'\circ m_T\in ca(\Sigma)$ and

\begin{equation*} (e'\circ T)(u)=\int_\Omega u(\omega)\,d(e'\circ m_T) ,\quad \mbox{for all}\ \ u\in B(\Sigma). \end{equation*}

Let $i:E\rightarrow E^{\prime\prime}$ stand for the canonical isometry, that is, $i(e)(e')=e'(e)$ for $e\in E$, $e'\in E'$ and $\|i(e)\|_{E^{\prime\prime}}=\|e\|_E$. Hence for each $e'\in E'$, we get

\begin{equation*} T^*(e') = e'\circ m_T =\Phi^{-1}(e'\circ T)= \sum^\infty_{n=1}\alpha_n\,i(e_n)(e')\,\lambda_n. \end{equation*}

This means that $T^*$ is a nuclear operator and by (2.5) we get $\|T^*\|_{nuc}\le|m_T|(\Omega)$.

Now, we shall show that

\begin{equation*} |m_T|(\Omega)\le\|T^*\|_{nuc}. \end{equation*}

Let ɛ > 0 be given. Since $T^*$ is a nuclear operator, there exist a bounded sequence $(e^{\prime\prime}_n)$ in E ^ʹʹ, a bounded sequence $(\lambda_n)$ in $ca(\Sigma)$ and $(\alpha_n)\in\ell^1$ so that

\begin{equation*} T^*(e')=\sum_{n=1}^\infty\alpha_n\,e^{\prime\prime}_n(e')\,\lambda_n\ \ \mbox{for}\ \ e'\in E' \end{equation*}

and

(2.6)\begin{equation} \sum_{n=1}^\infty|\alpha_n|\|e^{\prime\prime}_n\|_{E^{\prime\prime}}|\lambda_n|(\Omega)\le\|T^*\|_{nuc}+\varepsilon. \end{equation}

Then for $A\in\Sigma$, we obtain

\begin{equation*} (e'\circ m_T)(A)=T^*(e')(A)=\sum_{n=1}^\infty\alpha_n\,e^{\prime\prime}_n(e')\,\lambda_n(A). \end{equation*}

Moreover, by the Hahn-Banach theorem for every $A\in\Sigma$, there exists $e'_A\in E'$ with $\|e'_A\|_{E'}=1$ such that $\|m_T(A)\|_E=|(e'_A\circ m_T)(A)|$. Hence, if Π is a finite Σ-partition of Ω, then using (2.6) we have

\begin{equation*} \begin{array}{l} \displaystyle\sum_{A\in\Pi}\|m_T(A)\|_E=\sum_{A\in\Pi}|(e'_A\circ m_T)(A)|=\sum_{A\in\Pi}\left|\sum_{n=1}^\infty\alpha_n\,e^{\prime\prime}_n(e'_A)\,\lambda_n(A)\right|\\[6mm] \displaystyle\leq\sum_{A\in\Pi}\left(\sum^\infty_{n=1}|\alpha_n|\,|e^{\prime\prime}_n(e'_A)|\,|\lambda_n(A)|\right)\leq\sum^\infty_{n=1}\left(|\alpha_n|\|e^{\prime\prime}_n\|_{E^{\prime\prime}}\sum_{A\in\Pi}|\lambda_n(A)|\right)\\[6mm] \displaystyle\leq\sum^\infty_{n=1}|\alpha_n|\|e^{\prime\prime}_n\|_{E^{\prime\prime}}|\lambda_n|(\Omega)\le\|T^*\|_{nuc}+\varepsilon. \end{array} \end{equation*}

Since ɛ > 0 is arbitrary, we get $|m_T|(\Omega)\le\|T^*\|_{nuc}$ and finally $\|T^*\|_{nuc}=|m_T|(\Omega)=\|T\|_{nuc}$.

3. Traces of Bochner representable operators

Formulas for the traces of kernel operators on Banach function spaces (in particular, $L^p(\mu)$-spaces) have been the object of much study (see [Reference Grothendieck14], [Reference Brislawn2], [Reference Delgado4], [Reference Gohberg, Goldberg and Krupnik10], [Reference Pietsch22]).

Grothendieck [Reference Grothendieck13, Chap. I, p. 165] showed that the notion of ‘trace’ can be defined for nuclear operators in Banach spaces with the approximation property (see [Reference Pietsch22, 4.6.2, Lemma, pp. 210–211]).

Recall that a Banach space $(X,\|\cdot\|_X)$ has the approximation property if for every compact subset K of X and every ɛ > 0 there exists a bounded finite rank operator $S:X\rightarrow X$ such that $\|x-S(x)\|_X\leq\varepsilon$ for every $x\in K$ (see [Reference Ryan23, Chap. 4, p. 72], [Reference Diestel and Uhl7, Definition 1, p. 238]).

Note that the Banach space $B(\Sigma)$ has the approximation property. Assume first that $B(\Sigma)$ is the Banach lattice of all bounded Σ-measurable real functions on Ω. Since $(B(\Sigma),\|\cdot\|_\infty)$ is an AM-space with the unit $\mathbb{1}_\Omega$, due to the Kakutani-Bohnenblust-M. and S. Krein theorem (see [Reference Aliprantis and Burkinshaw1, Theorem 3.40]) $B(\Sigma)$ is lattice isometric to some C(K)-space for a unique (up to homeomorphism) compact Hausdorff space K in such a way that $\mathbb{1}_\Omega$ is identified with $\mathbb{1}_K$. This follows that $B(\Sigma)$ has the approximation property because C(K) has the approximation property (see [Reference Ryan23, Example 4.2]). For the Banach space $B(\Sigma)$ of complex-valued functions on Ω, one has to consider real and imaginary parts separate.

Assume that $T:B(\Sigma)\rightarrow B(\Sigma)$ is a nuclear operator, that is, there exist a bounded sequence $(\lambda_n)$ in $ba(\Sigma)$, a bounded sequence $(w_n)$ in $B(\Sigma)$ and $(\alpha_n)\in\ell^1$ so that

(3.1)\begin{equation} T=\sum^\infty_{n=1}\alpha_n\Phi_{\lambda_n}\otimes w_n \ \ \mbox{in} \ \ {\cal L}(B(\Sigma),B(\Sigma)). \end{equation}

Then the trace of T is given by

\begin{equation*} {\rm tr}\,T:=\sum^\infty_{n=1}\alpha_n\Phi_{\lambda_n}(w_n)=\sum^\infty_{n=1}\alpha_n\int_\Omega w_n(\omega)\,d\lambda_n, \end{equation*}

and it does not depend on the special choice of the nuclear representation (3.1) of T (see [Reference Gohberg, Goldberg and Krupnik10, Chap. 5, Theorem 1.2], [Reference Pietsch22, Lemma, pp. 210–211]).

From now on we assume that $(\Omega,{\cal T})$ is a compact Hausdorff space and ${\cal B} o$ denotes the σ-algebra of Borel sets in Ω. Then $C(\Omega)\subset B({\cal B} o)$.

Assume that a measure $\mu\in ca^+({\cal B} o)$ is strictly positive, that is, for all $U\in{\cal T}$ with $U\neq \emptyset$, $\mu(U) \gt 0$. Then $L^1(\mu,C(\Omega))\subset L^1(\mu,B({\cal B} o))$.

Corollary 3.1. Assume that $T:B({\cal B} o)\rightarrow B({\cal B} o)$ is a Bochner representable operator such that

\begin{equation*} T(u)=\int_\Omega u(\omega)\,f(\omega)\,d\mu,\quad \mbox{for all} \ u\in B({\cal B} o), \end{equation*}

where $f\in L^1(\mu, C(\Omega))$. Then T has a well-defined trace

\begin{equation*} {\rm tr}\,T=\int_\Omega f(\omega)(\omega)\,d\mu. \end{equation*}

Proof. Let $L^1(\mu)\,\hat{\otimes}\,C(\Omega)$ denote the projective tensor product of $L^1(\mu)$ and $C(\Omega)$, equipped with the completed norm π (see [Reference Diestel and Uhl7, p. 227], [Reference Ryan23, p. 17]). Note that for $z\in L^1(\mu)\,\hat{\otimes}\, C(\Omega)$, we have

\begin{equation*} \pi(z)=\inf\left\{\sum^\infty_{n=1}|\alpha_n|\,\|v_n\|_1\|w_n\|_\infty\right\}, \end{equation*}

where the infimum is taken over all sequences $(v_n)$ in $L^1(\mu)$ and $(w_n)$ in $C(\Omega)$ with $\lim_n\|v_n\|_1=0=\lim_n\|w_n\|_\infty$ and $(\alpha_n)\in\ell^1$ such that $z=\sum^\infty_{n=1}\alpha_n\,v_n\otimes w_n$ in π-norm (see [Reference Ryan23, Proposition 2.8, pp. 21–22]).

It is known that $L^1(\mu)\,\hat{\otimes}\, C(\Omega)$ is isometrically isomorphic to the Banach space $(L^1(\mu,C(\Omega)),\|\cdot\|_1)$ by the isometry J, defined by:

\begin{equation*} J(v\otimes w):=v\, (\cdot)\, w \ \ \mbox{for} \ v\in L^1(\mu), \ w\in C(\Omega) \end{equation*}

(see [Reference Diestel and Uhl7, Example 10, p. 228], [Reference Ryan23, Example 2.19, p. 29]). Then there exist sequences $(v_n)$ in $L^1(\mu)$ and $(w_n)$ in $C(\Omega)$ with $\lim_n\|v_n\|_1=0=\lim_n\|w_n\|_\infty$ and $(\alpha_n)\in\ell^1$ such that

\begin{equation*} J^{-1}(f)=\sum^\infty_{n=1}\alpha_n\,v_n\otimes w_n \ \ \mbox{in} \ \big(L^1(\mu)\,\hat{\otimes} \,C(\Omega), \pi\big). \end{equation*}

Thus it follows that

\begin{equation*} f=J\left(\sum^\infty_{n=1}\alpha_n v_n\otimes w_n\right)=\sum^\infty_{n=1} \alpha_n v_n(\cdot)\, w_n \ \ \mbox{in} \ \ L^1(\mu,C(\Omega)), \end{equation*}

and hence

\begin{equation*} T(u)=\sum^\infty_{n=1}\alpha_n\left(\int_\Omega u(\omega)\,v_n(\omega)\,d\mu\right) w_n,\quad \mbox{for all} \ \ u\in B(\Sigma). \end{equation*}

For $n\in\mathbb{N}$, let

\begin{equation*} \lambda_n(A):=\int_A v_n(\omega)\,d\mu,\quad \mbox{for all} \ \ A\in\Sigma. \end{equation*}

Note that $\lambda_n\in ca(\Sigma)$ and $|\lambda_n|(\Omega)=\|v_n\|_1$ and hence $\lim\lambda_n(A)=0$ for all $A\in\Sigma$. By the Nikodym convergence theorem (see [Reference Drewnowski9, Theorem 8.6]), the family $\{\lambda_n:n\in\mathbb{N}\}$ is uniformly countably additive.

Since $\Phi_{\lambda_n}(u)=\int_\Omega u(\omega)\,d\lambda_n=\int_\Omega u(\omega)\,v_n(\omega)\,d\mu$ for all $u\in B(\Sigma)$ (see [Reference Conway3, Theorem 8C, p. 380]), we get

\begin{equation*} T(u)=\sum^\infty_{n=1}\alpha_n\Phi_{\lambda_n}(u)w_n,\quad \mbox{for all} \ \ u\in B(\Sigma), \end{equation*}

that is,

\begin{equation*} T=\sum_{n=1}^\infty\alpha_n\Phi_{\lambda_n}\otimes w_n\ \ \mbox{in}\ \ {\cal L}(B(\Sigma),B(\Sigma)). \end{equation*}

Hence

\begin{equation*} {\rm tr}\,T=\sum_{n=1}^\infty\alpha_n\Phi_{\lambda_n}(w_n)=\sum_{n=1}^\infty\alpha_n\int_\Omega w_n(\omega)\,v_n(\omega)\,d\mu. \end{equation*}

For $n\in\mathbb{N}$, let $f_n=\sum_{i=1}^n\alpha_i\,v_i(\cdot)\,w_i$. Hence $\int_\Omega\|f(\omega)-f_n(\omega)\|_\infty\,d\mu\rightarrow 0$. Thus we get,

\begin{equation*} \begin{array}{l} \displaystyle \left|\int_\Omega f(\omega)(\omega)\,d\mu-\sum^n_{i=1}\alpha_i\int_\Omega v_i(\omega)\, w_i(\omega)\,d\mu\right|\\[5mm] \displaystyle \leq \int_\Omega\left|\left(f(\omega)(\omega)-\sum^n_{i=1}\alpha_i v_i(\omega)\,w_i(\omega)\right) \right|d\mu \leq \int_\Omega\|f(\omega)-f_n(\omega)\|_\infty \,d\mu. \end{array} \end{equation*}

Let $g\in L^1(\mu,C(\Omega))$ be another function representing T, that is,

\begin{equation*} T(u)(t)=\int_\Omega u(\omega)\,f(\omega)(t)\,d\mu(\omega)=\int_\Omega u(\omega)\,g(\omega)(t)\,d\mu(\omega)\ \ \mbox{for}\ \ u\in B({\cal B} o). \end{equation*}

Denote $h(\omega)(t):=f(\omega)(t)-g(\omega)(t)$ for $\omega,t\in\Omega$. Then for every $A\in{\cal B} o$ and $u=\mathbb{1}_A$ we obtain

\begin{equation*} \int_A h(\omega)(t)\,d\mu(\omega)=0 \ \ \mbox{for all}\ \ t\in\Omega. \end{equation*}

Hence for every $t\in\Omega$, $h(\cdot)(t)=0$ µ-a.e and it follows that

(3.2)\begin{equation} \int_\Omega\left(\int_\Omega |h(\omega)(t)|\,d\mu(\omega)\right)\,d\mu(t)=0. \end{equation}

We shall show that

\begin{equation*} \int_\Omega h(\omega)(\omega)\,d\mu(\omega)=0. \end{equation*}

For indirect proof suppose that $\left|\int_\Omega h(\omega)(\omega)\,d\mu(\omega)\right| \gt 0$. Then there exists $A\in{\cal B} o$, $\mu(A)\neq 0$ such that $h(\omega)(\omega) \gt 0$ or $h(\omega)(\omega) \lt 0 $ for $\omega\in A$. Without loss of generality, let $h(\omega)(\omega) \gt 0$ for $\omega\in A$. Since for $\omega\in\Omega$ we have $h(\omega)\in C(\Omega)$, then there exists a neighbourhood H_ω of $\omega\in A$ such that

\begin{equation*} h(\omega)(t) \gt 0\ \ \mbox{for every}\ \ t\in H_\omega. \end{equation*}

Since µ is strictly positive, then for every $\omega\in A$, $\mu(H_\omega) \gt 0$ and hence

\begin{equation*} \int_{H_\omega}h(\omega)(t)\,d\mu(t) \gt 0. \end{equation*}

Let $\omega_0\in A$ be given. Then, we have

\begin{equation*} \int_{\Omega}|h(\omega_0)(t)|\,d\mu(t)\ge\int_{\bigcup H_\omega}|h(\omega_0)(t)|\,d\mu(t)\ge\int_{H_{\omega_0}}|h(\omega_0)(t)|\,d\mu(t) \gt 0. \end{equation*}

Since ω ₀ is arbitrary, it follows that

\begin{equation*} \int_\Omega\left(\int_\Omega |h(\omega)(t)|\,d\mu(t)\right)\,d\mu(\omega) \gt 0 \end{equation*}

and, in view of Hille’s theorem (see [Reference Dinculeanu8, § 1, Theorem 36, p. 16]), this is in contradiction with (3.2). Hence we finally get

\begin{equation*} \int_\Omega h(\omega)(\omega)\,d\mu(\omega)=0. \end{equation*}

Thus this follows that the trace of T is well defined and ${\rm tr}\,T=\int_\Omega f(\omega)(\omega)\,d\mu$.

Grothendieck [Reference Grothendieck14] showed that if Ω is a compact Hausdorff space with a positive Borel measure µ on Ω and $k(\cdot,\cdot)\in C(\Omega\times\Omega)$, then the kernel operator $T_k:C(\Omega)\rightarrow C(\Omega)$ defined by:

\begin{equation*} T_k(u):=\int_\Omega u(\omega)\,k(\cdot,\omega)\,d\mu \ \ \mbox{for} \ \ u\in C(\Omega), \end{equation*}

is nuclear and has a well-defined trace ${\rm tr}\,T_k=\int_\Omega k(\omega,\omega)\,d\mu$ (see [Reference Grothendieck14], [Reference Pietsch22, 6.6.2, Theorem, p. 274]).

Now, we can extend this formula for the trace of kernel operators $T_k:B({\cal B} o)\rightarrow B({\cal B} o)$.

Let $k(\cdot,\cdot)\in C(\Omega\times\Omega)$. Hence for every $\omega\in\Omega$, $k(\cdot,\omega)\in C(\Omega)$. Let $C(\Omega, C(\Omega))$ denote the Banach space of all continuous functions $f:\Omega\rightarrow C(\Omega)$, equipped with the uniform norm $\|\cdot\|_\infty$.

Assume that $\mu\in ca({\cal B} o)^+$. Let ${\cal L}^\infty(\mu, C(\Omega))$ denote the space of all µ-measurable functions $g:\Omega\rightarrow C(\Omega)$ such that $\mu-{\rm ess }\sup\|g(\omega)\|_\infty \lt \infty$. In view of the Pettis measurability theorem (see [Reference Diestel and Uhl7, Theorem 2, p. 42]), we have

(3.3)\begin{equation} C(\Omega, C(\Omega))\subset{\cal L}^\infty(\mu, C(\Omega)), \end{equation}

and the space ${\cal L}^\infty(\mu, C(\Omega))$ can be embedded in the space $L^1(\mu, C(\Omega))$ such that with each function from ${\cal L}^\infty(\mu, C(\Omega))$ is associated its µ-equivalence class in $L^1(\mu, C(\Omega))$.

It is well known (see [Reference Pietsch22, 6.1.4, p. 243]) that the function:

\begin{equation*} f:\Omega\ni\omega\mapsto k(\cdot,\omega)\in C(\Omega), \end{equation*}

is bounded and continuous. Then in view of (3.3), $f\in{\cal L}^\infty(\mu, C(\Omega))$. Hence its µ-equivalence class belongs to $L^1(\mu,C(\Omega))$. Thus it follows that one can define the kernel operator $T_k: B({\cal B} o)\rightarrow B({\cal B} o)$ by

\begin{equation*} T_k(u):=\int_\Omega u(\omega)\,k(\cdot,\omega)\,d\mu,\quad \mbox{for all}\ \ u\in B({\cal B} o). \end{equation*}

For $t\in\Omega$, let $\Phi_t(u)=u(t)$ for all $u\in B({\cal B} o)$. Then $\Phi_t\in C(\Omega)'$ and using Hille’s theorem, for all $u\in B({\cal B} o)$, $t\in\Omega$, we get

\begin{equation*} T_k(u)(t)=\int_\Omega u(\omega)\,\Phi_t(k(\cdot,\omega))\,d\mu=\int_\Omega u(\omega)\,k(t,\omega)\,d\mu. \end{equation*}

As a consequence of Theorem 2.2 and Corollary 3.1, we get

Corollary 3.2. The kernel operator $T_k:B({\cal B} o)\rightarrow B({\cal B} o)$ is nuclear σ-smooth and

\begin{equation*} {\rm tr}\,T_k=\int_\Omega k(\omega,\omega)\,d\mu. \end{equation*}