Analysis Problem of the Day 72

Today’s problem appeared as Problem 2 on the UCLA Spring 2024 Qualifying Exam:

Problem 2. Let $K \subset \mathbb{R}$ be a compact set with positive Lebesgue measure, i.e. $|K|>0,$ and define

$K_n:=\{x: d(x,K) \leq \frac{1}{n}\}, \quad \mu_n(A):= \frac{|A \cap K_n|}{|K_n|}.$

Suppose $\mu_n \to \mu$ in the weak- $^*$ topology. Show that $|\text{supp}(\mu)|=|K|.$

Solution: Recall that the support of a measure is the complement of the union of null open sets with respect to the measure. Moreover, recall that weak- $^*$ convergence of measures implies that $\mu_n(A) \to \mu(A)$ for all measurable sets $A$ by applying the definition of weak- $^*$ convergence to the characteristic function $\chi_A.$ We first note that $\text{supp}(\mu_n) \subset K_n.$ Indeed, $K_n$ is closed as the preimage of a closed set under a continuous function, and so for any $x \not \in K_n,$ some open neighborhood of $x$ does not intersect $K_n,$ and thus is $\mu_n$ -null according to the definition of $\mu_n.$ We now show that this implies that $\text{supp}(\mu) \subset K.$ Indeed, if not, there exists $x \not \in K$ and an open neighborhood $U \ni x$ a positive distance away from $K$ (since $K$ is compact and $\{x\}$ is compact, the two sets are a positive distance apart, as they are disjoint) with positive $\mu$ -measure. Then, $\mu_n(U) \to \mu(U)>0,$ so $\mu_n(U)>0$ for large enough $n,$ i.e. $U \cap \text{supp}(\mu_n) \not = \varnothing.$ But $K_n \cap U = \varnothing$ for large enough $n$ since $d(K_n,\overline{U})>0$ for $\sup_{a \in K_n, b \in K} d(K_n,K) \leq \frac{d(\overline{U},K)}{2},$ which is a contradiction. Thus, $\text{supp}(\mu) \subset K.$

Finally, we claim that $|K \setminus \text{supp}(\mu)|=0.$ If not, there exists $A \subset K$ such that $|A|>0$ and $A \cap \text{supp}(\mu)=\varnothing.$ Note that $\bigcap_{n=1}^\infty K_n =K,$ so by continuity of measure, $|K_n| \to |K|.$ But

$\mu_n(A) = \frac{|A \cap K_n|}{|K_n|} = \frac{|A|}{|K_n|} \to \frac{|A|}{|K|}=\mu(A)>0,$

which contradicts that $A \cap \text{supp}(\mu) = \varnothing.$ Thus, $\text{supp}(\mu) \subset K$ and $|K \setminus \text{supp}(\mu)|=0,$ i.e. $|K|=|\text{supp}(\mu)|,$ as desired.

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