I need some help with the proof of the following.

Let $(X,\mathcal{A},\mu)$ be a measure space with $\mu$ being $\sigma$-finite, $\mu^*$ be the outer measure given by the formula $\mu^*(E)=\inf\{\sum_{n}\mu(A_n): E\subset\bigcup A_n, (A_n)\subset\mathcal{A}\}$ and $\mathcal{M}$ the $\sigma$-algebra of the $\mu^*$-measurable sets. Also let $\mathcal{A}_{\mu}=\{A\subset X: \exists E,F\in\mathcal{A},$ with $E\subset A\subset F$ and $ \mu(F\setminus E)=0\}$ and $\overline{\mu}:\mathcal{A}_\mu\to [0,+\infty]$ given by $\overline{\mu}(E)=\sup\{\mu(B): B\in\mathcal{A}, B\subset E\}.$

Then $\mathcal{A}_\mu=\mathcal{M}$ and $\overline{\mu}=\mu^*\vert_{\mathcal{M}}$.

Okay, so it is known that the measure space $(X,\mathcal{A}_\mu, \overline\mu)$ is the completion of $(X,\mathcal{A},\mu)$ and by the facts that $\mu^*\vert_\mathcal{A}=\mu$ and $\mathcal{A}\subset\mathcal{M}$, we have that $\mathcal{A}_\mu\subset\mathcal{M}$ and that $\mu^*\vert_{\mathcal{A}_\mu}=\overline\mu$. So in order to prove the statement above, it would be enough to show that $\mathcal{M}\subset\mathcal{A}_\mu$. I'm trying to prove that if $A\in\mathcal{M}$ and $\mu^*(A)<\infty$ then $A\in\mathcal{A}_{\mu}$ but I'm stuck. My progress is the following:

For each $n\in\mathbb{N}$ there exists a sequence $(A_k^{(n)})\subset\mathcal{A}$ such that $\sum_{k}\mu(A_k^{(n)})<\mu^*(A)+\frac{1}{n}$. Let $A_n=\bigcup_{k}A_k^{(n)}\in\mathcal{A}$. We have that $\mu(A_n)\leq\sum_{k}\mu(A_k^{(n)})<\mu^*(A)+\frac{1}{n}$. Finally let $F=\bigcap_{n} A_n$. Then $\mu(F)\leq\mu(A_n)<\mu^*(A)+\frac{1}{n}$ for all $n$. Taking limits we have that $\mu(F)\leq\mu^*(A)$, but $A\subset F$, so $\mu^*(A)\leq\mu^*(F)=\mu(F)$ so we found the first desirable set. But what about the other? I'm stuck here and I can't seem to be able to use the sigma-finiteness. Any help?

EDIT: Maybe considering the fact that the sigma-algebra $\mathcal{A}_1=\{A\cup E: A\in\mathcal{A}, E\subset F,$ where $F\in\mathcal{A}, \mu(F)=0\}$ is equal to $\mathcal{A}_\mu$ helps. It is easy to show this equality.


Denote by $(X_n)_{n \in \mathbb{N}} \subseteq \mathcal{A}$ a sequence of increasing sets, $X_n \uparrow X$, such that $\mu(X_n)< \infty$ for all $n \in \mathbb{N}$. You have already shown that

$$B \in \mathcal{M} \implies \exists F \in \mathcal{A}, F \supseteq B: \mu(F) = \mu^*(B). \tag{1}$$

Roughly speaking, the idea is to use $(1)$ for $B:=A^c \in \mathcal{M}$ and then define $E:=F^c$; if $\mu$ is a finite measure, you can easily verify that the so-defined set has all desired properties. For $\sigma$-finite measures the reasoning is somewhat more complicated.

Fix $A \in \mathcal{M}$. Applying $(1)$ for $B:=A^c \cap X_n \in \mathcal{M}$ we find that there exists a set $H_n \in \mathcal{A}$, $H_n \supseteq A^c \cap X_n$, such that

$$\mu(H_n) = \mu^*(A^c \cap X_n)<\infty.$$

Note that $E_n := X_n \backslash H_n \in \mathcal{A}$ satisfies $E_n \subseteq A \cap X_n$ and

$$\mu(E_n) = \mu(X_n)-\mu(H_n) = \mu^*(X_n)-\mu^*(A^c \cap X_n) = \mu^*(X_n \backslash A^c) = \mu^*(X_n \cap A)$$

where we have used in the penultimate step that $A^c \in \mathcal{M}$. Combining this with the fact that $X_n \cap A^c \in \mathcal{M}$, we find that

$$\mu^*((X_n \cap A) \backslash E_n)=0. \tag{2}$$

Now set $E:= \bigcup_{n \geq 1} E_n \in \mathcal{A}$, then $E \subseteq A$ and, by the monotonicity and subadditivity of $\mu^*$,

$$\begin{align*} \mu^*(A \backslash E) &= \mu^* \left( \left( \bigcup_{n \geq 1} A \cap X_n \right) \backslash \left( \bigcup_{k \geq 1} E_k \right) \right) \\ &\leq \mu^* \left( \bigcup_{n \geq 1} (A \cap X_n) \backslash E_n \right) \\ &\leq \sum_{n \geq 1} \mu^*((A \cap X_n) \backslash E_n) \\ &\stackrel{(2)}{=} 0.\end{align*}$$


$$\mu^*(A) = \mu^*(E) = \mu(E).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.