Analysis Problem of the Day 27

Today’s problem is Problem 1 on the UCLA Spring 2023 Analysis Qual:

Problem 1: Let $P$ be the set of all Borel probability measures on $[0,1]$ and let $m$ be the Lebesgue measure.

a) Show that $\mu \in P$ satisfies $\mu \ll m$ if and only if for all $\epsilon>0$ there exists a $\delta>0$ such that for all $f \in C([0,1]), 0 \leq f \leq 1,$ $\int f dm < \delta$ implies $\int f d\mu < \epsilon.$

b) Show that $A=\{\mu \in P: \mu \ll m\}$ is a Borel subset of $P$ for the weak-* topology.

Solution: a) Recall that $\mu \ll m,$ i.e. $\mu$ is absolutely continuous with respect to $m,$ if and only if $m(E) = 0 \implies \mu(E)=0.$ Additionally, recall the following equivalent characterization reminiscent of continuity: $\mu \ll m$ if and only if for all $\epsilon>0,$ there exists a $\delta>0$ such that $\mu(E)<\epsilon$ whenever $m(E)<\delta.$ First, suppose $\mu \ll m.$ Note that for any $f \in C([0,1]), 0 \leq f \leq 1,$ $m_f:= \int f dm, \mu_f := \int f d\mu$ are Borel measures on $[0,1].$ Then, the statement of the problem is equivalent to showing that $\mu \ll m \Longleftrightarrow \mu_f \ll m_f$ for all $f$ as above. Indeed, fix $f$ and suppose $m_f(E) = 0.$ Then, $\int f dm = 0,$ i.e. $m(E \cap \{f >0\})=0.$ By absolutely continuity, this implies $\mu(E \cap \{f>0\})=0,$ so $\int_E f d\mu = \mu_f(E)=0.$

Conversely, by density of continuous functions in $L^1,$ let $f_n \to \chi_E$ in $L^1(\mu+m),$ i.e. both in $L^1(\mu)$ and $L^1(m).$ If $\{f_n\}$ is unbounded in $L^\infty,$ one may replace $f_n$ with $\max(f_n,M)$ and still have $f_n \in C([0,1]), f_n \to \chi_E$ in $L^1(\mu+m).$ Consequently, we may assume $|f_n| \leq M$ for all $n.$ Moreover, notice that if $m(E) < \delta,$ then $\int |f_n| dm \leq M \delta$ for all $n.$ Then,

$\mu(E) = \int \chi_E d\mu \leq \|\chi_E-f_n\|_{L^1(\mu)}+\int |f_n| d\mu< 2\epsilon$

whenever $\|\chi_E-f_n\|_{L^1(\mu)}<\epsilon$ and $\int |f_n| d\mu<\epsilon$ (by uniform absolute continuity of the $\mu_{f_n}$ and the fact that $\int |f_n|dm < M \delta$ ). Thus, the statements are equivalent.

b) Since $[0,1]$ is a Polish space (that is, a separable completely metrizable topological space), by Prokhorov’s theorem, the weak-* topology on the unit ball in $C([0,1])^*$ is completely metrizable. In particular, closedness is equal to sequential closedness. I claim that for fixed $\epsilon, \delta>0,$ the set

$C_{\epsilon,\delta}=\left\{\mu \in P: \int f d\mu \leq \epsilon \text{ for all }f\text{ such that} \int f dm \leq \delta, 0 \leq f \leq 1, f\in C([0,1])\right\}$

is weak-* closed. Indeed, if $\mu_n \in A$ , $\mu_n \rightharpoonup \mu,$ then $\int f d\mu_n \to \int fd\mu \leq \epsilon.$ Then, using a), we conclude that

$A = \bigcap_{\epsilon_n=\frac{1}{n}} \bigcup_{\delta_m = \frac{1}{m}} C_{\epsilon_n,\delta_m},$

which is a Borel set as a countable union/intersection of weak-* closed sets. Thus, $A$ is Borel for the weak-* topology.

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