Show that $\inf {\{m^∗(A \cap B): A, B \ m^∗ \text{–measurable}, A \supseteq E, B \supseteq X\backslash E}\} > 0$.

Question

The following is an exercise from Bruckner's Real Analysis:

Let $m^∗$ be an outer measure on a set $X$ , and suppose that $E ⊂ X$ is not $m^∗$–measurable. Show that $\inf {\{m^∗(A \cap B): A, B \ m^∗ \text{–measurable}, A \supseteq E, B \supseteq X\backslash E}\} > 0$.

I have no idea how to approach the problem at all . A useful hint also would be great, thanks!

Answer 1

Let $m^∗$ be an outer measure on a set $X$ , and suppose that $E \subset X$ is not $m^∗$–measurable. Show that $\inf {\{m^∗(A \cap B): A, B \ m^∗ \text{–measurable}, A \supseteq E, B \supseteq X\setminus E}\} > 0$.

Proof: As usual, we write $E^c$ to mean $X\setminus E$. Suppose $\inf {\{m^∗(A \cap B): A, B \ m^∗ \text{–measurable}, A \supseteq E, B \supseteq E^c}\} = 0$.

Then there are two sequences $A_n$ and $B_n$, $m^*$-measurable such that $A_n \supseteq E, B_n \supseteq E^c$ and $m^∗(A_n \cap B_n)$ converges to $0$ as $n \to \infty$.

Let $A=\bigcap_n A_n$ and $B=\bigcap_n B_n$. we have

1. $A$ is $m^*$-measurable and $E \subseteq A$

2. $B$ is $m^*$-measurable and $E^c \subseteq B$

3. Since, for all $n$, $A\cap B \subseteq A_n \cap B_n$, and $m^∗(A_n \cap B_n)$ converges to $0$ as $n \to \infty$, we have that $m^∗(A \cap B)=0$

Note that, from item 1 above, we have $A^c \subseteq E^c \subseteq B$.

Now, given any $C \subseteq X$, we have

\begin{align*} m^*(C) &\leq m^*(C\cap E) + m^*(C\cap E^c) \leq \\ & \leq m^*(C\cap A) + m^*(C\cap B) \leq & \\ & \leq m^*(C\cap A) + m^*(C\cap (B \cap A^c)) + m^*(C\cap (B \cap A)) = \\ & = m^*(C\cap A) + m^*(C\cap (B \cap A^c)) = \\ & = m^*(C\cap A) + m^*(C\cap A^c) = \\ & = m^*(C) \end{align*} So, $ m^*(C) = m^*(C\cap E) + m^*(C\cap E^c) $. So $E$ is $m^*$-measurable. Contradiction. So, we have $\inf {\{m^∗(A \cap B): A, B \ m^∗ \text{–measurable}, A \supseteq E, B \supseteq E^c}\} > 0$.