Elegant Problems in Measure Theory

Footnotes

1. Let us approximate $x$ by dyadic rationals $q_n(x):=2^{-n}\lfloor 2^n x\rfloor$, so $q_n(x)\xrightarrow[n\to\infty]{}x$.

   Since the floor function outputs integers, $q_n(x)$ can only take the form $k2^{-n}$ for some $k\in\mathbb{Z}$.

   If $\lfloor 2^n x\rfloor=k$, then

   $$k\leq 2^n x \lt k+1,$$

   hence

   $$\frac{k}{2^n}\leq x \lt \frac{k+1}{2^n}.$$

   Therefore, for fixed $n$,

   $$\lbrace x\in\mathbb{R}:q_n(x)=k2^{-n}\rbrace = \left[\frac{k}{2^n},\frac{k+1}{2^n}\right).$$

   Hence, on this interval, $q_n(x)$ is constant and equals $k2^{-n}$.

   In the strip

   $$\left[\frac{k}{2^n},\frac{k+1}{2^n}\right)\times\mathbb{R},$$

   the condition $f(q_n(x),y)>\alpha$ becomes $f(k2^{-n},y)>\alpha$.

   For fixed $n,k,\alpha$, define

   $$A_{n,k,\alpha}:=\lbrace y\in\mathbb{R}:f(k2^{-n},y)>\alpha\rbrace.$$

   Since $f(x,\cdot)$ is Lebesgue measurable for every $x$, the set $A_{n,k,\alpha}\subseteq\mathbb{R}$ is Lebesgue measurable.

   Define $g_n(x,y):=f(q_n(x),y)$. Then

   $$\lbrace (x,y)\in\mathbb{R}^2:g_n(x,y)>\alpha\rbrace = \bigcup_{k\in\mathbb{Z}} \left[\frac{k}{2^n},\frac{k+1}{2^n}\right) \times A_{n,k,\alpha}.$$

   Each interval $\left[\frac{k}{2^n},\frac{k+1}{2^n}\right)$ is Borel measurable, and each $A_{n,k,\alpha}$ is Lebesgue measurable. Hence each product

   $$\left[\frac{k}{2^n},\frac{k+1}{2^n}\right) \times A_{n,k,\alpha}$$

   is Lebesgue measurable in $\mathbb{R}^2$.

   Thus $\lbrace (x,y)\in\mathbb{R}^2:g_n(x,y)>\alpha\rbrace$ is Lebesgue measurable, since it is a countable union of Lebesgue measurable sets.

   Since $\alpha\in\mathbb{R}$ is arbitrary, $g_n$ is Lebesgue measurable.

   By continuity of $f(\cdot,y)$,

   $$g_n(x,y)=f(q_n(x),y)\xrightarrow[n\to\infty]{}f(x,y).$$

   Hence $g_n$ converges pointwise to $f$. Since the pointwise limit of Lebesgue measurable functions is Lebesgue measurable, $f$ is Lebesgue measurable. ↩

2. Recall that the topology on $X$ is discrete if and only if every singleton set in $X$ is open.

   For the first direction, suppose the counting measure $\mu$ is a Radon measure on $X$.

   Let $x\in X$ be arbitrary and consider the singleton set $A=\lbrace x\rbrace$, then $\mu(A)=1$ by definition of counting measure, since $|A|=1$. By the outer regularity property of a Radon measure, $\inf \lbrace \mu(G): x\in G, G\text{ open}\rbrace=1$.

   Since the counting measure $\mu$ takes values in $\lbrace 0,1,2,\dots, \infty\rbrace$, the infimum must hit 1 exactly, so $\exists G_0$ open with $x\in G_0$ and $\mu(G_0)=|G_0|=1$. Hence, we must have $G_0=\lbrace x\rbrace$, so $\lbrace x\rbrace$ is open. Since $x\in X$ was arbitrary, we know that every singleton set in $X$ is open, hence the topology on $X$ is discrete.

   This proves that $\mu$ is a Radon measure $\implies$ the topology on $X$ is discrete.

   For the second direction, suppose the topology on $X$ is discrete.

   Under the counting measure, every subset of the space $X$ is measurable, since the $\sigma$-algebra is the power set $\mathcal{P}(X)$. Since the Borel set is a subcollection of the power set, Borel measurability follows immediately. Now it suffices to check inner regularity, outer regularity and local finiteness.

   Let us note that under the discrete topology, every set is open, since one can take a ball of radius $r<1$ around any point $x\in X$, and then $B_r(x)=\lbrace x\rbrace$. Hence, the complement of every set is open, hence every set is closed. So every subset of $X$ is clopen under the discrete topology.

   Also, note that for a set $K$, the set $\bigcup_{x\in K} \lbrace x\rbrace$ forms an open cover of $K$, so for $K$ to be compact there must be a finite subcover. This is only possible if $K$ is finite. Hence, under the discrete topology, compact $\Leftrightarrow$ finite.

   Given arbitrary $x\in X$, we know that $U=\lbrace x\rbrace$ is an open neighbourhood of $x$ with $\mu(U)=|U|=|\lbrace x\rbrace|=1<\infty$. This proves local finiteness.

   For any $A\subseteq X$, we have $\inf \lbrace \mu(G): A \subseteq G, G\text{ open}\rbrace= \mu(G_0)$ where $G_0=\bigcup_{a\in A} \lbrace a\rbrace$. Hence, $\mu(G_0)=|A|=\mu(A)$. This proves outer regularity.

   For any $A\subseteq X$, we want to find $\sup \lbrace \mu(K): K \subseteq A, K\text{ compact}\rbrace$. If $A$ is finite then $A$ is compact, hence we pick $K=A$, then $\mu(K)=\mu(A)$. If $A$ is infinite then we can pick $n$ elements from $A$ to form $K_n$, so $K_n$ is finite hence it is compact, and we let $n\to \infty$ so we have $\sup_n(\mu(K_n))=\infty=\mu(A)$. This proves inner regularity.

   This proves that the topology on $X$ is discrete $\implies$ $\mu$ is a Radon measure. ↩

3. We claim that if $A\in \Sigma$, then there are only two possibilities: either $A\subseteq \mathbb{Q}$, or $A=B\cup(\mathbb{R}\setminus \mathbb{Q})$ with $B\subseteq \mathbb{Q}$.

   Consider the collection $\mathcal{F}=\lbrace A\subseteq \mathbb{R}: A\subseteq \mathbb{Q} \text{ or } \mathbb{R}\setminus \mathbb{Q} \subseteq A\rbrace$. Let us show that $\Sigma=\mathcal{F}$.

   First, we want to show that $\mathcal{F}\subseteq \Sigma$. It suffices to show that $\mathcal{F}$ is a $\sigma$-algebra and that $\mathcal{F}$ contains all subsets of $\mathbb{Q}$.

   Let $A\in \mathcal{F}$. If $A\subseteq \mathbb{Q}$ then $A^c=\mathbb{R}\setminus A=(\mathbb{R}\setminus \mathbb{Q})\cup(\mathbb{Q}\setminus A)$, so $\mathbb{R}\setminus \mathbb{Q}\subseteq A^c$ and $A^c\in \mathcal{F}$.

   If $\mathbb{R}\setminus \mathbb{Q}\subseteq A$, then $A\cap A^c=\emptyset$ implies $(\mathbb{R}\setminus \mathbb{Q})\cap A^c=\emptyset$, so $A^c\subseteq \mathbb{Q}$ and $A^c\in \mathcal{F}$.

   Let $\bigcup_{n=1}^\infty A_n$ be a countable union of elements of $\mathcal{F}$. If there exists $n_0$ such that $\mathbb{R}\setminus \mathbb{Q}\subseteq A_{n_0}$, then $\mathbb{R}\setminus \mathbb{Q}\subseteq \bigcup_{n=1}^\infty A_n$, so $\bigcup_{n=1}^\infty A_n\in \mathcal{F}$.

   Otherwise, for all $n$, $A_n\subseteq \mathbb{Q}$, so $\bigcup_{n=1}^\infty A_n\subseteq \mathbb{Q}$ and $\bigcup_{n=1}^\infty A_n\in \mathcal{F}$.

   Any arbitrary $B\subseteq \mathbb{Q}$ is in $\mathcal{F}$ by definition, so $\mathcal{F}$ contains all subsets of $\mathbb{Q}$.

   Second, we want to show that $\Sigma\subseteq \mathcal{F}$. Since $\mathcal{F}$ is a $\sigma$-algebra that contains all subsets of $\mathbb{Q}$, and $\Sigma$ is the smallest $\sigma$-algebra that contains all subsets of $\mathbb{Q}$, it is clear that $\Sigma\subseteq \mathcal{F}$.

   Thus, $\Sigma=\lbrace A\subseteq \mathbb{R}: A\subseteq \mathbb{Q} \text{ or } \mathbb{R}\setminus \mathbb{Q} \subseteq A\rbrace$.

   Consider $f:(\mathbb{R},\Sigma)\to(\mathbb{R},\mathcal{B})$ defined as $f(x)=x^2$. Then $B=\lbrace 2\rbrace\in \mathcal{B}$ is a Borel set, but $f^{-1}(\lbrace 2\rbrace)=\lbrace \sqrt{2},-\sqrt{2}\rbrace\notin \Sigma$, since it contains some but not all irrationals. So $f(x)=x^2$ is a continuous function, hence a Borel function, but it is not $\Sigma$-measurable.

   Conversely, there does not exist a $\Sigma$-measurable function which is not Borel. Indeed, $\Sigma\subseteq \mathcal{B}$. If $M\in \Sigma$ and $M\subseteq \mathbb{Q}$, then $M$ is countable, so it is a countable union of singleton Borel sets and hence Borel. If $\mathbb{R}\setminus \mathbb{Q}\subseteq M$, then $M=(\mathbb{R}\setminus \mathbb{Q})\cup \tilde{M}$ with $\tilde{M}\subseteq \mathbb{Q}$, and both pieces are Borel. Hence $M\in \mathcal{B}$. ↩

4. Suppose for sake of contradiction that there exists such an $X$.

   Let $x\in X$. Take the selective intersection $\bigcap$ of all sets $Y \subset \Sigma$, i.e., include $Y$ in the intersection if $x\in Y$ and include $Y^c$ in the intersection if $x\notin Y$. This intersection gives us the smallest set in $\Sigma$ containing $x$, call it $A_x$. We will refer to this as the 'atom' of $x$.

   Consider $x,y\in X$ with $x\neq y$. Suppose $z\in A_x\cap A_y$.

   Then, $z\in A_x$ and $A_x\in \Sigma$ implies that $A_z\subseteq A_x$ (since $A_z$ is the smallest set in $\Sigma$ containing $z$). By the same logic, $A_z\subseteq A_y$.

   Now suppose $x\notin A_z$, then $x\in A_z^c\in \Sigma$, hence $A_x \subseteq A_z^c$ because $A_z^c$ is a particular set containing $x$ and $A_x$ is the intersection of all such sets. But we know $A_z \subseteq A_x$ so this is a contradiction, hence $x\in A_z$. Thus, $A_x\subseteq A_z$ (since $A_x$ is the smallest set in $\Sigma$ containing $x$). So $A_x=A_z$ and by the same reasoning, $A_y=A_z$.

   So $A_x\cap A_y\neq \emptyset \implies A_x=A_y.$

   So the distinct atoms are pairwise disjoint and their union is all of $X$. Every set in $\Sigma$ must be a union of atoms. So if there are $n$ distinct atoms, we can form $nCn+nC(n-1)+\dots+nC1=2^n$ sets in $\Sigma$. So if there are countably infinite distinct atoms, we can form uncountably infinite sets in $\Sigma$.

   If there are finitely many atoms then $\Sigma$ is countably infinite. If there are countably infinitely many atoms then $\Sigma$ is uncountably infinite. In neither case is a countably infinite $\Sigma$ possible. ↩

5. Suppose for sake of contradiction that there exists such a measure.

   Since $X$ is an infinite dimensional Banach space, the closed unit ball in $X$ (call it $U$) is not compact. $U$ is a closed subset of $X$ and $X$ is complete, hence $U$ is complete. We know that 'compact' $\Leftrightarrow$ 'complete and totally bounded'.

   Hence, $U$ is not totally bounded, so $\exists \delta>0$ such that $U$ cannot be covered with finitely many balls of radius $\delta$.

   Pick any $e_1\in U$. Since $U$ is not covered by $B(e_1, \delta)$, pick $e_2\in U\setminus B(e_1,\delta)$. Inductively, pick $e_{n+1} \in U\setminus \bigcup_{k=1}^n B(e_k,\delta)$.

   This is a bounded sequence of points ${e_i}_{i\in \mathbb{N}}$ in the closed unit ball such that $\|e_i-e_j\|_X \geq \delta>0$ for all $i\neq j$ and $\|e_i\|\leq 1$ for all $i$.

   Consider the balls $B_i := B(e_i,\frac{\delta}{3})$. Clearly, $B_i\cap B_j=\emptyset$ for $i\neq j$ (since $\frac{2\delta}{3}<\delta$).

   By translation invariance, $\mu(B_i)=k$ for all $i$, where $k$ is some finite positive constant. This is because $B_j=B(e_j,\frac{\delta}{3})=B(e_i,\frac{\delta}{3}) + (e_j-e_i) = B_i + (e_j-e_i)$, hence $\mu(B_j)=\mu(B_i)$.

   Consider a finite set of such balls, then by finite additivity, we have $\mu(\bigcup_{i=1}^n B_i) = \sum_{i=1}^n \mu(B_i) = nk$.

   But $\bigcup_{i=1}^n B_i \subset B(0, 1+\frac{\delta}{3})$, hence $\mu(\bigcup_{i=1}^n B_i) = nk \le \mu(B(0, 1+\frac{\delta}{3}))$.

   But since this holds for all $n$, this forces $\mu(B(0, 1+\frac{\delta}{3})) = \infty$ which is a contradiction. ↩