Proof. For $N,Q\in \mathbb N$ , with $Q\ge 2$ , define

$$ \begin{align*} U_{N,Q} = \bigcup_{q\ge Q} \bigcup_{p\in\mathbb Z} \bigg( \frac pq-\frac1{q^N}, \frac pq+\frac1{q^N} \bigg). \end{align*} $$

Each $U_{N,Q}$ is open. It is also dense in $\mathbb R$ , since it contains all rational numbers $p/q$ with $q\ge Q$ , and these are dense in $\mathbb R$ .

We claim that

$$ \begin{align*} \mathcal L = \bigg( \bigcap_{N=1}^{\infty} \bigcap_{Q=2}^{\infty} U_{N,Q} \bigg) \cap(\mathbb R\setminus\mathbb Q). \end{align*} $$

Indeed, the inclusion from right to left is immediate. Conversely, if $x\in \mathcal L$ , then x is irrational. Given N and Q, choose $M>N$ . By the Liouville property, there exists $p/q$ such that

$$ \begin{align*} \bigg|x-\frac{p}{q}\bigg|<q^{-M}. \end{align*} $$

For fixed Q, the denominators q occurring in such approximations cannot remain bounded as $M\to \infty $ , since x is irrational. Hence, for some such approximation, $q\ge Q$ , and then

$$ \begin{align*} \bigg|x-\frac{p}{q}\bigg|<q^{-M}\le q^{-N}. \end{align*} $$

Thus, $x\in U_{N,Q}$ . The first factor is a countable intersection of open dense sets. Also, $\mathbb R\setminus \mathbb Q$ is a dense $G_\delta $ , since

$$ \begin{align*} \mathbb R\setminus\mathbb Q = \bigcap_{r\in\mathbb Q}(\mathbb R\setminus\{r\}). \end{align*} $$

Therefore, $\mathcal L$ is a dense $G_\delta $ subset of $\mathbb R$ .

Finally, the classical Liouville theorem implies that every Liouville number is transcendental. Since the set of algebraic real numbers is countable, this does not by itself give measure zero. The standard Borel–Cantelli argument gives measure zero: for fixed $N>2$ , the total length of the intervals with $q\ge Q$ over any bounded interval is bounded by a constant multiple of

$$ \begin{align*} \sum_{q\ge Q} q\cdot q^{-N} = \sum_{q\ge Q} q^{1-N}, \end{align*} $$

which tends to $0$ as $Q\to \infty $ . Hence, the set of real numbers satisfying the Liouville inequalities for this fixed $N>2$ has measure zero on every bounded interval. Since $\mathcal L$ is contained in that set, $\mathcal L$ has Lebesgue measure zero.

Proof. Let $f\colon (0,\infty )\to (0,\infty )$ be the function $f(x)=x^x=\exp (x\log x)$ . The derivative of f is $f'(x)=x^x(\log x+1)$ , and thus f is strictly decreasing on $(0,e^{-1})$ and strictly increasing on $(e^{-1},\infty )$ , with a unique global minimum $f(e^{-1})=e^{-1/e}$ . Consequently, for each $\alpha \in \mathbb {R},$ the level set $F_\alpha = f^{-1}(\{\alpha \})\subset (0,\infty )$ has cardinality $|F_\alpha |\leq 2$ .

Consider the set

$$ \begin{align*} A := \bigcup_{\alpha\in \overline{\mathbb{Q}}\cap\mathbb{R}} F_\alpha = \{x> 0 : x^x \text{ is algebraic}\}. \end{align*} $$

Since each $F_\alpha $ is finite and $ \overline {\mathbb {Q}}\cap \mathbb {R}$ is countable, A is countable. Hence, A is meagre and has Lebesgue measure $0$ . Therefore, its complement $\mathcal {T}=(0,\infty )\setminus A$ is comeagre (equivalently, a dense $G_\delta $ ) and has full Lebesgue measure.

Proof. By Theorem 2.2, $\mathcal T$ is a dense $G_\delta $ subset of $(0,\infty )$ . By Lemma 2.1, $\mathcal L\cap (0,\infty )$ is also a dense $G_\delta $ . Since the intersection of two dense $G_\delta $ subsets of a Baire space is again a dense $G_\delta $ , the result follows.

Proof. Let $f:(0,\infty )\to (0,\infty )$ be given by $f(x)=x^x$ . The function f is continuous, and $f'(x)=x^x(\log x+1)$ . Hence, $f'(x)=0$ only at $x=e^{-1}$ . Since $\mathcal L$ is a $G_\delta $ subset of $\mathbb R$ and f is continuous, the preimage $ \mathcal G=f^{-1}(\mathcal L)$ is a $G_\delta $ subset of $(0,\infty )$ .

We next prove density. Let $I\subset (0,\infty )$ be a nonempty open interval. Since f is continuous and nonconstant on I, the image $f(I)$ contains a nonempty open interval J. Since $\mathcal L$ is dense in $\mathbb R$ , we have $J\cap \mathcal L\neq \varnothing $ . Choose $y\in J\cap \mathcal L$ . Since $J\subset f(I)$ , there exists $x\in I$ such that $f(x)=y$ . Therefore $x\in I\cap \mathcal G$ . Hence, $I\cap \mathcal G\neq \varnothing $ , so $\mathcal G$ is dense in $(0,\infty )$ .

It remains to prove that $\mathcal G$ has Lebesgue measure zero. Let $c=e^{-1}$ . Since ${f'(x)\neq 0}$ for every $x\neq c$ , every compact interval $K\subset (0,\infty )\setminus \{c\}$ has the property that ${|f'(x)|\ge m_K>0}$ for all $x\in K$ . Since K is an interval not containing c, the derivative $f'$ has constant sign on K. Moreover, by the mean value theorem,

$$ \begin{align*} |f(x)-f(y)|\ge m_K |x-y| \quad (x,y\in K). \end{align*} $$

Therefore, $f|_K$ is injective and its inverse branch is Lipschitz. In particular, the inverse branch $(f|_K)^{-1}$ is Lipschitz on $f(K)$ .

Since $\mathcal L$ has Lebesgue measure zero, the set $\mathcal L\cap f(K)$ has measure zero. Since $f|_K$ is injective and its inverse branch is Lipschitz, the image of every null set under $(f|_K)^{-1}$ is a null set. Since $\mathcal L\cap f(K)$ has measure zero,

$$ \begin{align*} (f|_K)^{-1}(\mathcal L\cap f(K)) = K\cap f^{-1}(\mathcal L) \end{align*} $$

also has measure zero.

Now, $(0,\infty )\setminus \{c\}$ is a countable union of compact intervals of the above type. For example,

$$ \begin{align*} (0,\infty)\setminus\{c\} = \bigcup_{n=1}^{\infty} \bigg[\frac1n,c-\frac1n\bigg] \cup \bigcup_{n=1}^{\infty} \bigg[c+\frac1n,n\bigg], \end{align*} $$

where empty intervals are ignored. Therefore, $f^{-1}(\mathcal L)\setminus \{c\}$ has measure zero. Adding the single point c does not change the measure, and hence

$$ \begin{align*} \mu(\mathcal G)=\mu(f^{-1}(\mathcal L))=0. \end{align*} $$

Thus, $\mathcal G$ is a dense $G_\delta $ subset of $(0,\infty )$ and has Lebesgue measure zero.

Proof. By Theorem 2.4, $\mathcal G$ is a dense $G_\delta $ subset of $(0,\infty )$ . By Lemma 2.1, $\mathcal L\cap (0,\infty )$ is also a dense $G_\delta $ subset of $(0,\infty )$ . Hence, $\mathcal G\cap \mathcal L$ is the intersection of two dense $G_\delta $ subsets of the Baire space $(0,\infty )$ and is therefore again a dense $G_\delta $ .