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...


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 $(*)$.

  • $\begingroup$ Thanks! So you don't even need to make a distinction between E having infinite outer measure or not. $\endgroup$ – Axesilo Apr 5 '12 at 19:48
  • $\begingroup$ then the set F would be dependent on the epsilon wouldn't it? $\endgroup$ – Mathcho Sep 10 '17 at 11:59

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.