# Outer Measure exercise

This comes from an exercise from Real Analysis by Folland.

Let $$\mathcal{A}\subset P(X)$$ be an algebra, $$\mathcal{A}_\sigma$$ the collection of countable unions of sets in $$\mathcal{A}$$, and $$\mathcal{A}_{\sigma\delta}$$ the collection of countable intersections of sets in $$\mathcal{A}_\sigma$$. Let $$\mu_{0}$$ be a premeasure on $$\mathcal{A}$$ and $$\mu^*$$ the induced outer measure.

a.) For any $$E\subset X$$ and $$\epsilon > 0$$ there exists $$A\in \mathcal{A}_\sigma$$ with $$E\subset A$$ and $$\mu^*(A) \leq \mu^*(E) + \epsilon$$.

b.) If $$\mu^*(E) < \infty$$, then $$E$$ is $$\mu^*$$-measurable if and only if there exists $$B\in \mathcal{A}_{\sigma\delta}$$ with $$E\subset B$$ and $$\mu^*(B\setminus E) = 0$$

proof of a.) was done in an earlier post: Outer measure problem

proof of b.) Using part a.) $$\forall \epsilon_i$$ there exists $$A_i\in \mathcal{A}_\sigma$$ such that $$E\subset A_i$$. Let $$B = \bigcap_{i=1}^{n}A_i$$ then $$B\in\mathcal{A}_{\sigma\delta}$$. Now, we need to show that $$\mu^*(B\cap E^c) = 0$$. From the definition of outer measure we know that: $$\mu^*(B\cap E^c) = \inf\left\{\sum_{1}^{\infty}\mu^*(E_j):E_j\in\mathcal A \ \ \text{and} \ \ (B\cap E^c)\subset \bigcup_{1}^{\infty}E_j\right\}$$ From part a.) we know that $$\mu^*(A)\leq \mu^*(E) + \epsilon$$. I am not sure if this fact helps and I am stuck where to go from here, any suggestions would be great. Also, any hints on how to prove the converse would be great as well.

• What exactly is $\mathcal{A}\subset \mathcal{A}$ supposed to mean?
– zoli
Commented Aug 19, 2015 at 17:28
• Sorry, that was a mis-type, see edit Commented Aug 19, 2015 at 17:29
• Hi @Wolfy , is there a solution for the other parts of the question? Commented Nov 1, 2020 at 10:45
• @user726608 There is a link for the answer. Commented Nov 10, 2020 at 6:16

A hint for the forward case:

Since by assumption $E$ is $\mu^{*}$-measurable we have that for all $F \subset X$ we have:

$\mu^{*}(F) = \mu^{*}(F \cap E) + \mu^{*}(F\cap E^{c})$

Since this holds for all $F \subset X$ it must also hold for $B$ as you constructed above. Now use the fact that $E \subset B$ and part (a) to rearrange the above and show that $0 \leq \mu^{*}(B\setminus E)< \varepsilon$.

A hint for the reverse case:

First if there exists such a set $B$ show that $B$ is $\mu^{*}$ measurable. Then you get that for any $F \subset X$:

$\mu^{*}(F) = \mu^{*}(F \cap B) + \mu^{*}(F\cap B^{c})$

Now try to show that:

$\mu^{*}(F \cap B) \geq \mu^{*}(F\cap B\cap E) + \mu^{*}(F\cap B \cap E^{c})$

and

$\mu^{*}(F \cap B^{c}) = \mu^{*}(F\cap B^{c} \cap E^{c} ) + \mu^{*}(F\cap B^{c} \cap E)$

Finally note that:

$F \cap E = (F\cap B\cap E)\cup(F\cap B^{c} \cap E)$

and

$F\cap E^{c} = (F\cap B\cap E^{c})\cup(F\cap B^{c} \cap E^{c})$

Combining all of this information with subadditivity of $\mu^{*}$ it is possible to show that $\mu^{*}(F) \geq \mu^{*}(F\cap E) + \mu^{*}(F \cap E^{c})$ which is sufficient to prove $E$ is $\mu^{*}$ measurable.

• In the forward case: How can $\mu^*(B\setminus(E)) \leq \epsilon$ be shown? And in the other case: why is $\mu^*(F)$ is finite (as it's necessary according to the theory)? Commented Oct 31, 2020 at 15:38
• can you explain it? Commented Nov 1, 2020 at 10:40