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My question has to do with the very last paragraph of the top answer to this question. In this paragraph, the author chooses sets $G_1$ and $G_2$ in advantageous ways. How are we allowed to choose these sets with the properties specified in the answer?

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Andrew
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1 Answers1

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First recall some notation from the other thread:

Given a measure space $(X, \Sigma, \mu)$, we can define an outer measure $\mu^\ast$ as follows. For an arbitrary $A \subseteq X$, set $$ \mu^\ast(A) = \inf \{\mu(E) : E \in \Sigma, E \supseteq A\} $$

This infimum is in fact a minimum when $\mu^\ast(A)$ is finite:

Claim 1. For each $A \subseteq X$ there is $G \in \Sigma$ with $G \supseteq A$ and $\mu(G) = \mu^\ast(A)$.

Proof. If $\mu^\ast(A) = \infty$, we can take $G = X$, so suppose $\mu^\ast(A) < \infty$. For every $n \in \mathbb{N}$ choose $G_n \in \Sigma$ with $G_n \supseteq A$ and $\mu(G_n) < \mu^\ast(A) + \frac{1}{n}$. Then $G = \bigcap_{n = 1}^\infty G_n \supseteq A$ has measure $\mu(G) = \mu^\ast(A)$.

As a slight refinement of the above:

Claim 2. Let $A \subseteq X$ be arbitrary and let $K \in \Sigma$ satisfy $K \supseteq A$. Then there is $H \in \Sigma$ that has $\mu(H) = \mu^\ast(A)$ and $A \subseteq H \subseteq K$.

Proof. If $\mu^\ast(A) = \infty$, take $H = K$. If $\mu^\ast(A) < \infty$, use the notation of the previous proof, put $H_n = G_n \cap K$ and $H = \bigcap_{n=1}^\infty H_n = G \cap K$.

With claim 2 in hand, it is now easy to find the desired sets $G_1$ and $G_2$:

In the answer, $F$ and $E_n$ are given, where $E_n \in \Sigma$ has finite measure.

It is claimed that with $F_n = E_n \cap F$, it is possible to choose $G_1 \in \Sigma$ with $F_n \subseteq G_1 \subseteq E_n$ and $\mu(G_1) = \mu^\ast(E_n \cap F_n)$.

To see this, apply claim 2 with $A = F_n$ and $K = E_n$, and find that $G_1 := H$ has the desired properties: $F_n \subseteq G_1 \subseteq E_n$ and $\mu(G_1) = \mu^\ast(F_n) = \mu^\ast(F_n \cap E_n)$.

Similarly for the existence $G_2$: set $A = E_n \setminus F_n$ and $K = E_n$ apply claim 2 and find that $G_2 = H$ has the desired properties.

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