Analysis Problem of the Day 42

Today’s problem appeared as Problem 2 on the UCLA Spring 2022 Analysis Qual:

Problem 2: Recall that a function $f: \mathbb{R} \to \mathbb{R}$ is called 1-Lipschitz if $|f(x)-f(y)| \leq |x-y|$ for all $x,y \in \mathbb{R}.$ Let $\mathcal{F}$ be an arbitrary family of 1-Lipschitz functions $f: \mathbb{R} \to \mathbb{R},$ and let $L_f:=\{(x,y) \in \mathbb{R}^2: f(x) \leq y\}.$

a) Show that $\bigcup_{f \in \mathcal{F}} L_f$ is Lebesgue measurable.

b) Show that $\bigcup_{f \in \mathcal{F}} L_f$ need not be Borel measurable.

Solution: a) Notice that $\bigcup_f L_f$ is the union of the hypographs of all $f \in \mathcal{F}.$ In particular, for fixed $x,$ $y$ belongs to the union if it is less than $f(x)$ for some function $f \in \mathcal{F}.$ Motivated by this, we note that

$|f(x)-f(x')| \leq |x-x'| \implies |\sup_{f \in \mathcal{F}} f(x)-\sup_{f \in \mathcal{F}} f(x')| \leq |x-x'|,$

where $g=\sup f: \mathbb{R} \to \overline{\mathbb{R}}$ can be thought of as an extended real-valued function. Since it is unclear whether the supremum is actually attained at each point (since we do not have strict inequality), note that $\bigcup_f L_f$ has to take the form $\bigcup_f L_f = N \cup \{(x,y) \in \mathbb{R}^2: y < g(x)\},$ where $N \subseteq g(\mathbb{R}).$ The latter set is Borel as it is the preimage $h^{-1}((0,\infty))$ of the continuous and therefore measurable function $h(x,y) = g(x)-y,$ and we claim the former is a null set. By countability, it suffices to prove $g([0,1])$ is null. But indeed, noting $N \subset C_\epsilon:=\bigcup_{\|f-g\|_\infty \leq \epsilon} f([0,1])$ for any $\epsilon >0,$ sending $\epsilon \to 0$ yields $\mu(N)=0,$ as desired. Thus, the union $\bigcup_{f \in \mathcal{F}} L_f$ is Lebesgue measurable.

b) Note that if $N$ is a non-Borel measurable subset of $\mathbb{R},$ (say, $N$ is the Vitali set), then we may construct a family $\mathcal{F}$ such that $g(x)=0,$ where $g$ is as above, and $\bigcup_f L_f = B \cup i(N)$ is the disjoint union of a Borel measurable set $B=\{(x,y) \in \mathbb{R}^2: y<0\}$ and $i(N):=\{(x,0): x \in N\}.$ Since the inclusion map $i: \mathbb{R} \to \mathbb{R}^2, i(x)=(x,0)$ is continuous, it follows that $i(N)$ is not Borel measurable as a subset of $\mathbb{R}^2$ either, as otherwise so would be its preimage $N.$ Then, $(B \cup i(N)) \cap B^c = i(N),$ which is a contradiction since the Borel $\sigma$ -algebra is closed under complements and intersections and $N$ is not Borel measurable. Finally, to construct such a family $\mathcal{F},$ it suffices to take 1-Lipschitz functions of the form $p(x)=-|x|$ and let

$\mathcal{F}:=\{p(x-a)-b: (a \in N \text{ and } b \leq 0) \text{ or } (a \not \in N \text{ and } b<0)\}.$

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