# If $\mu^*$ is an outer measure induced by a premeasure and $E$ is locally measurable, then why is $E$ measurable?

This is essentially Exercise 1.21 in Folland's Real Analysis, which states the following: If $\mu^*$ is an outer measure induced by a premeasure and $\overline{\mu}$ is the restriction of $\mu^*$ to the $\mu^*$-measurable sets, then $\overline{\mu}$ is saturated.

Definition: Folland says a measure $\overline{\mu}$ on a space $(X, \mathcal{M})$ is saturated if every locally measurable set is measurable, where a set $E$ is locally measurable if and only if $E \cap A$ is measurable for every $A \in \mathcal{M}$ with $\overline{\mu}(A) < \infty$.

I'm having trouble showing that when $E$ is a locally measurable set with $\mu^*(E) = \infty$, then $E$ is measurable. (The finite case is not hard.)

Here's what I have so far. Write $\mu_0$ for the premeasure, $\mathcal{A} \subset \mathcal{P}(X)$ for the algebra on which $\mu_0$ is defined, and $\mathcal{M}$ for the collection of all $\mu^*$-measurable sets. Also, let $\mathcal{A}_{\sigma}$ be the collection of countable unions of sets in $\mathcal{A}$. The hint is to use an earlier exercise: for any $\varepsilon > 0$, there is $A \in \mathcal{A}_{\sigma}$ with $E \subset A$ and $\mu^*(A) \leq \mu^*(E) + \varepsilon$. So I obtain $E = \bigcup_{j=1}^{\infty} E \cap A_j$, where each $A_j \in \mathcal{A}$. Then I want to use the locally measurable property, but it may be the case that $\mu_0(A_j) = \infty$ for some $j_0$.

Any ideas? We don't have any assumption that $\overline{\mu}$ is $\sigma$-finite, for example...

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Assume $E$ is a locally measurable set. It suffices to show that $$\mu^*(F)\geq \mu^*(F\cap E)+\mu^*(F\cap E^c)\quad(*)$$ for all $F\subset X$ with $\mu^*(F)<\infty$. So let $F\subset X$ be such and let $\epsilon>0$. Using the earlier exercise, we find an $A\in\mathcal{A}_\sigma\subset\mathcal{M}$ with $F\subset A$ and $\mu^*(A)\leq\mu^*(F)+\epsilon$ (thus $\mu^*(A)<\infty$). Now $E\cap A\in\mathcal{M}$, since $E$ is locally measurable and $A\in\mathcal{M}$ is such that $\mu^*(A)<\infty$. It follows that $$\mu^*(A)=\mu^*(A\cap(E\cap A))+\mu^*(A\cap(E\cap A)^c)=\mu^*(A\cap E)+\mu^*(A\cap E^c).$$ Since $F\subset A$, $$\mu^*(A\cap E)+\mu^*(A\cap E^c)\geq \mu^*(F\cap E)+\mu^*(F\cap E^c).$$ All in all, we have $$\mu^*(F)+\epsilon\geq \mu^*(A) \geq \mu^*(F\cap E)+\mu^*(F\cap E^c),$$ and since $\epsilon>0$ was arbitrary, we get $(*)$.

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Thanks! So you don't even need to make a distinction between E having infinite outer measure or not. – DA29731 Apr 5 '12 at 19:48

2

down vote

accepted

"Assume E is a locally measurable set. It suffices to show that μ ∗ (F)≥μ ∗ (F∩E)+μ ∗ (F∩E c )(∗) for all F⊂X with μ ∗ (F)<∞ .... "

The above proof tries to use the definition of outer measurability. In this, however, the sets F are not restricted to sets of bounded measure. If the generating premeasure is non-semifinite, the proof will therefore fail.

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