外勒贝格测度命题求证：勒贝格可测集相关测度等式证明

Hey there, let's break this proof down clearly— I’ve had my fair share of head-scratching with outer measure and measurable set existence problems, so I know this can feel tricky at first. Let's start by restating the problem to make sure we're on the same page:

Given a Lebesgue measurable set $E$, and any subset $A \subset E$, we need to find a Lebesgue measurable set $B$ such that:

1. $E \setminus A \subset B \subset E$
2. $\mu(B) = \mu^(E \setminus A)$ (where $\mu$ is Lebesgue measure, $\mu^$ is Lebesgue outer measure)

Step 1: Use Outer Measure Regularity to Get a $G_\delta$ Set #

First, recall a key property of Lebesgue outer measure: for any set $S \subset \mathbb{R}^n$, there exists a $G_\delta$-set (countable intersection of open sets) $G$ such that $S \subset G$ and $\mu(G) = \mu^*(S)$.

Let's set $S = E \setminus A$. Applying this regularity property, we get a $G_\delta$-set $G$ with:

• $E \setminus A \subset G$
• $\mu(G) = \mu^*(E \setminus A)$

Since $G_\delta$-sets are always Lebesgue measurable (all Borel sets are measurable), $G$ is measurable.

Step 2: Construct the Desired Set $B$ #

Now, we need to "trim" $G$ to fit inside $E$. Define:
$$B = G \cap E$$

Let's verify the core conditions one by one:

1. $E \setminus A \subset B$

We know $E \setminus A \subset G$ (from Step 1) and $E \setminus A \subset E$ (by definition of set difference). Therefore, $E \setminus A$ is a subset of both $G$ and $E$, so it's contained in their intersection: $E \setminus A \subset G \cap E = B$.

2. $B \subset E$

This is immediate from the definition of $B$ as the intersection of $G$ and $E$.

3. $B$ is Lebesgue measurable

We know $E$ is measurable (given) and $G$ is measurable (as a $G_\delta$-set). The intersection of two measurable sets is always measurable, so $B$ is measurable.

Step 3: Verify the Measure Equality $\mu(B) = \mu^*(E \setminus A)$ #

We'll use two foundational facts here:

• For measurable sets, outer measure equals measure: $\mu^*(B) = \mu(B)$
• Measure is monotonic: if $X \subset Y$, then $\mu(X) \leq \mu(Y)$

First, since $E \setminus A \subset B$, outer measure monotonicity gives:
$$\mu^(E \setminus A) \leq \mu^(B) = \mu(B)$$

Second, since $B = G \cap E \subset G$, measure monotonicity gives:
$$\mu(B) \leq \mu(G)$$

But from Step 1, $\mu(G) = \mu^(E \setminus A)$. Combining these inequalities:
$$\mu^(E \setminus A) \leq \mu(B) \leq \mu^*(E \setminus A)$$

This forces $\mu(B) = \mu^*(E \setminus A)$, which is exactly what we needed to show.

Quick Intuition Check #

The key idea here is using the regularity of Lebesgue outer measure to get a measurable set that "approximates" $E \setminus A$ from the outside, then intersecting with $E$ to keep it within our original measurable set $E$. Since $E$ is measurable, this intersection preserves measurability, and the measure equality holds thanks to the monotonicity of measure and outer measure.

内容的提问来源于stack exchange，提问作者Lukaszmat