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--- math105-s22:hw:hw2 [2022/01/27 14:58]
pzhou [Lemma 0]
+++ math105-s22:hw:hw2 [2022/02/03 23:00] (current)
pzhou [Lemma 3]
@@ Line 28: / Line 28: @@
 Hint: This is a hard one. First prove that $A$ can be written as countable union of bounded closed subsets, then suffice to prove the claim that any bounded closed (hence compact) subset $A$ is measurable.
+Hint 2: Given $A$ a closed bounded subset of $\R^n$, for any $\epsilon>0$, we have some open set $U$ with $m^*(U) < m^*(A) + \epsilon$. Then we just need to show that $m^*(U \RM A) < \epsilon$. Note $U\RM A $ is open. Claim: for any closed subset $K \in U \RM A$, we have $m^*(K) + m^*(A) = m^*(A \cup K) \leq m^*(U)$, hence $m^*(K) < \epsilon$. The key point is to show that there exists an increasing sequence of closed subsets $K_n \In U \RM A$, such that $\lim m^*(K_n) = m^*(U \RM A)$. Try to construct $K_n$ as finite union of almost disjoint closed boxes, e.g., use the dyadic subdivision trick.
 What does a closed set look like? Say, the cantor set in $[0,1]$?

Lecture Notes