Hint on a measure theory question

Question

Let $(X, \mathcal{A}, \mu)$ be a measure space, and define $\mu^\bullet: \mathcal{A} \to [0, +\infty]$ by

$$ \mu^\bullet (A) = \sup \left\{ \mu(B) : B \subseteq A, B \in \mathcal{A}, \mu(B) < + \infty \right\} $$ I'm trying to show that $\mu^\bullet$ is countably additive. One thing is clear: if $A_n$ are disjoint sets in $\mathcal{A}$, then if $B \subseteq \cup_n A_n$, and $B_n = B \cap A_n$, then $$ \mu(B) = \sum_n \mu(B_n) \leq \sum_n \mu^\bullet(A_n), $$ thus $\mu^\bullet(\cup_n A_n) \leq \sum_n \mu^\bullet(A_n)$. This is the easy direction (I think).

Could I get a small hint on the reverse inequality? (I don't want the solution...)

Answer 1

Hint: fix $\epsilon>0$ and for each $n$ take $B_n$ with $\mu(B_n)+\epsilon/2^n>\mu(A_n)$, then try to use the fact that $\sum_n\epsilon/2^n=\epsilon$.

Answer 2

Hint:
1. If $\mu^\bullet(A_n)=\infty$ for some $n$, show that $\mu^\bullet(\bigcup_n A_n)=\infty$.
2. If for all $n$, $\mu^\bullet(A_n)<\infty$, we can find $B_n\subseteq A_n$ such that $\mu(B_n)>\mu^\bullet(A_n)-\epsilon/2^n$. Let $B=\bigcup_n B_n\subseteq \bigcup_n A_n$.
3. If $\mu(B)=\infty$, try to use $\color{blue}B_N=\bigcup_{n=1}^N B_n$ instead, and show that $\mu^\bullet(\bigcup_n A_n)=\infty$.