Doubt about a problem in Measure Theory

Question

I was trying to solve the following problem.

Let $(X,\mu, \mathcal{F})$ be a measure space. For $E \subseteq X,$ we define $$\mu_{*}(E)=\sup \{ \mu (A) : A \in \mathcal{F} \text{ and } A \subseteq E \},$$ and $$\mu^{*}(E)=\inf \{ \mu (B) : B \in \mathcal{F} \text{ and } E \subseteq B \}.$$ Show that $\mathcal{F}_\mu = \{ E \subseteq X : \mu_{*}(E)=\mu^{*} (E) \}.$

Remark: $\mathcal{F}_\mu := \{ A\cup N: A\in \mathcal{F}, \quad N\subseteq M \in \mathcal{F}, \quad \mu (M)=0 \}.$

I have the inclusion ($\subseteq$), but I can't figure it out how to do the other inclusion. Could someone help me with that inclusion? Thanks in advance.

Answer 1

We can write $F_{\mu}$ as follows: $$F_{\mu} = \{E \subseteq X: \exists A,B \in \mathcal{F}, A \subseteq E \subseteq B, \mu(A) = \mu(B)\}. $$ That is, the members of $F_{\mu}$ are exactly those sets that are "squashed" between two measurable sets of the same measure.

Given some set $E$ with $\mu_*(E) = \mu^*(E)$, can you find a measurable set contained in $E$ with measure $\mu_*(E)$, and a measurable set containing $E$ also with measure $E$? (Hint: if $\mu^*(E) = \mu_*(E)$ is finite, then we can take a sequence of increasing sets contained in $E$ which have limit measure $\mu_*(E)$, and similiarly for a sequence of decreasing sets containing $E$).

Answer 2

The reverse is only valid if $\mu^{*} (E)$ is finite. Here is a detailed proof.

Suppose $ E \subseteq X $ and $\mu_{*}(E)=\mu^{*} (E)$ and $\mu^{*} (E)$ is finite. Since $$\mu_{*}(E)=\sup \{ \mu (A) : A \in \mathcal{F} \text{ and } A \subseteq E \},$$ Let $\{A_n\}_n$ be a sequence of sets in $\mathcal{F}$, such that $$ \mu_{*}(E)-\frac{1}{n+1} \leqslant \mu(A_n) \leqslant \mu_{*}(E)$$ Note that, $\bigcup_n A_n \in \mathcal{F}$ and $\bigcup_n A_n \subseteq E$, so $ \mu(\bigcup_n A_n ) \leqslant \mu_{*}(E)$. So for all $n$, we have $$ \mu_{*}(E)-\frac{1}{n+1} \leqslant \mu(A_n) \leqslant \mu \left (\bigcup_n A_n \right ) \leqslant \mu_{*}(E)$$ So $\mu(\bigcup_n A_n ) = \mu_{*}(E)$.

Since $$\mu^{*}(E)=\inf \{ \mu (B) : B \in \mathcal{F} \text{ and } E \subseteq B \},$$ Let $\{B_n\}_n$ be a sequence of sets in $\mathcal{F}$, such that $$ \mu^{*}(E) \leqslant \mu(B_n) \leqslant \mu^{*}(E) +\frac{1}{n+1}$$ Note that, $\bigcap_n B_n \in \mathcal{F}$ and $E \subseteq \bigcap_n B_n $, so $ \mu^{*}(E) \leqslant \mu(\bigcap_n B_n )$. So for all $n$, we have $$ \mu^{*}(E) \leqslant \mu \left (\bigcap_n B_n \right ) \leqslant \mu(B_n) \leqslant \mu^{*}(E) + \frac{1}{n+1} $$ So $\mu(\bigcap_n B_n ) = \mu^{*}(E)$.

Let $A=\bigcup_n A_n $ and $B=\bigcap_n B_n$. Then we have $A, B\in \mathcal{F}$, $ A\subseteq E \subseteq B$ and $\mu(A) = \mu_{*}(E)= \mu^{*}(E) = \mu(B)$.

Since $\mu^{*}(E)$ is finite, we have $$ \mu(B\setminus A)=\mu(B) -\mu(A) = 0$$

Take $N=E\setminus A$ and $M= B\setminus A$ and we have

And we have $E=A\cup N$ , where $N \subseteq M$ and $\mu(M)=0$

Remark: Even if $(X,\mathcal{F}, \mu)$ is a $\sigma$-finite a measure space, the "reverse inclusion" does not hold for $E$ with $\mu^{*} (E)$ infinite.

Consider $X= \mathbb{R}$, $\mathcal{F}$ the Lebesgue $\sigma$-algebra and $\mu$ the Lebesgue measure. Let $V$ be a non-mensurable subset of $[0,1]$

Let $E=(-\infty, 0) \cup V$. Note that $E$ is not measurable, but it is easy to see that $\mu_{*}(E)= \mu^{*}(E)=\infty$.

If we could write $E$ as $A\cup N: A\in \mathcal{F}, \quad N\subseteq M \in \mathcal{F}, \quad \mu (M)=0 $, then since the Lebesgue measure is complete, $N$ would be measurable and so $E$ would be measurable. Contradiction.