Is A Measurable if the Inner and Outer Measures are Equal?

[ryo0071]

Let $$\lambda(A)$$ denote the measure of $$A$$ and let $$\lambda^{*}(A)$$ denote the outer measure of $$A$$ and let $$\lambda_{*}(A)$$ denote the inner measure of $$A$$

Okay so the question is as follows:

Suppose that $$A \cup B$$ is measurable and that
$$\lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) < \infty$$

Prove that $$A$$ and $$B$$ are measurable.

So I know that $$\lambda^{*}(A) < \infty$$ and $$\lambda^{*}(B) < \infty$$ otherwise it would contradict $$\lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) < \infty$$ (since if one or both was infinity the sum would be infinity)

So I think it would be enough to show that inner measure is equal to the outer measure. We know that $$\lambda_{*}(A) \leq \lambda^{*}(A)$$ for all sets $$A$$ so I would need to show $$\lambda^{*}(A) \leq \lambda_{*}(A)$$ but I'm not sure how to go about showing this. So first off am I on the right track? Second is how would I go about showing $$\lambda^{*}(A) \leq \lambda_{*}(A)$$ if I am on the right track.

Any help is appreciated.

 

[Opalg]

Let $$\lambda(A)$$ denote the measure of $$A$$ and let $$\lambda^{*}(A)$$ denote the outer measure of $$A$$ and let $$\lambda_{*}(A)$$ denote the inner measure of $$A$$

Okay so the question is as follows:

Suppose that $$A \cup B$$ is measurable and that
$$\lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) < \infty$$

Prove that $$A$$ and $$B$$ are measurable.

So I know that $$\lambda^{*}(A) < \infty$$ and $$\lambda^{*}(B) < \infty$$ otherwise it would contradict $$\lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) < \infty$$ (since if one or both was infinity the sum would be infinity)

So I think it would be enough to show that inner measure is equal to the outer measure. We know that $$\lambda_{*}(A) \leq \lambda^{*}(A)$$ for all sets $$A$$ so I would need to show $$\lambda^{*}(A) \leq \lambda_{*}(A)$$ but I'm not sure how to go about showing this. So first off am I on the right track? Second is how would I go about showing $$\lambda^{*}(A) \leq \lambda_{*}(A)$$ if I am on the right track.

There are several different approaches to measure theory, so it would help to know how you are defining inner and outer measure. Also, what space do $A$ and $B$ lie in – $\mathbb{R}$, $\mathbb{R}^n$, a topological space, or a general measure space?

 

[ryo0071]

Sorry about that. We have that $$A, B \subset \mathbb{R}^n$$.

We developed measure has follows:

Let $$a, b \in \mathbb{R}^n$$. A special rectangle is $$I = \{x \in \mathbb{R}^n | a_i \leq x_i \leq b_i $$ for $$1 \leq i \leq n\}$$
The measure of $$I$$ is $$\lambda(I) = (b_1 - a_1)\cdots(b_n - a_n)$$

We define a special polygon $$P$$ to be a finite union of non overlapping special rectangles. So if $$P = \bigcup_{n = 0}^N I_n$$ with $$I_n$$ special rectangles, then the measure of P is $$\lambda(P) = \sum_{n = 0}^N \lambda(I_n)$$

We defined the measure of an open set $$G$$ to be:
$$\lambda(G) = sup\{\lambda(P)|P \subset G$$ and $$P$$ is a special polygon$$\}$$
The measure of a compact set $$K$$ as:
$$\lambda(K) = inf\{\lambda(G)|K \subset G$$ and $$G$$ is an open set$$\}$$

With the outer measure defined as:
$$\lambda^{*}(A) = inf\{\lambda(G)|A \subset G$$ and $$G$$ is an open set$$\}$$
And the inner measure:
$$\lambda_{*}(A) = sup\{\lambda(K)|K \subset A$$ and $$K$$ is a compact set$$\}$$

I think that should cover everything.

(Also I'm new at writing math on the computer, how would I make it so the limits on the union and the sum were on the side instead of on the top and bottom?)

 

[Opalg]

Let $$\lambda(A)$$ denote the measure of $$A$$ and let $$\lambda^{*}(A)$$ denote the outer measure of $$A$$ and let $$\lambda_{*}(A)$$ denote the inner measure of $$A$$

Okay so the question is as follows:

Suppose that $$A \cup B$$ is measurable and that
$$\lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) < \infty$$

Prove that $$A$$ and $$B$$ are measurable.

My first thought on this was that $A$ and $B$ must be almost disjoint (by which I mean that $A\cap B$ should be a null set). If not, then surely it would be true that $\lambda^{*}(A\cup B) < \lambda^{*}(A) + \lambda^{*}(B)$?

I have not checked this through carefully, but here is a rough outline of why I think $A$ should be measurable. I will write $\sim$ to mean "approximately equals", leaving you to translate into a proper proof with epsilons.

Let $U$ be an open set containing $A$, with $\lambda(U) \sim \lambda^{*}(A)$, and let $V$ be an open set containing $B$, with $\lambda(V) \sim \lambda^{*}(B)$. Then $U\cup V \supseteq A\cup B$ and $\lambda(U\cup V) \sim \lambda(A\cup B)$.

Now let $K$ be a compact set contained in $A\cup B$, with $\lambda(K) \sim \lambda(A\cup B)$. Define $L = K-V$. Then $L$ is a compact set contained in $A$. I think that it should be possible to show that $\lambda(L) \sim \lambda(U)$. If so, then $A$ is trapped between $L$ and $U$, whose measures are almost the same. Therefore $A$ is measurable.

(Also I'm new at writing math on the computer, how would I make it so the limits on the union and the sum were on the side instead of on the top and bottom?)

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