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

[ryo0071]

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

Okay so the question is as follows:

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

Prove that \(\displaystyle A\) and \(\displaystyle B\) are measurable.

So I know that \(\displaystyle \lambda^{*}(A) < \infty\) and \(\displaystyle \lambda^{*}(B) < \infty\) otherwise it would contradict \(\displaystyle \lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) < \infty\) (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 \(\displaystyle \lambda_{*}(A) \leq \lambda^{*}(A)\) for all sets \(\displaystyle A\) so I would need to show \(\displaystyle \lambda^{*}(A) \leq \lambda_{*}(A)\) 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 \(\displaystyle \lambda^{*}(A) \leq \lambda_{*}(A)\) if I am on the right track.

Any help is appreciated.

 

[Opalg]

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

Okay so the question is as follows:

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

Prove that \(\displaystyle A\) and \(\displaystyle B\) are measurable.

So I know that \(\displaystyle \lambda^{*}(A) < \infty\) and \(\displaystyle \lambda^{*}(B) < \infty\) otherwise it would contradict \(\displaystyle \lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) < \infty\) (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 \(\displaystyle \lambda_{*}(A) \leq \lambda^{*}(A)\) for all sets \(\displaystyle A\) so I would need to show \(\displaystyle \lambda^{*}(A) \leq \lambda_{*}(A)\) 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 \(\displaystyle \lambda^{*}(A) \leq \lambda_{*}(A)\) 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 \(\displaystyle A, B \subset \mathbb{R}^n\).

We developed measure has follows:

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

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

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

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

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

Okay so the question is as follows:

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

Prove that \(\displaystyle A\) and \(\displaystyle B\) 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?)

To produce textstyle LaTeX, use \$ ... \$ tags instead of [MATH] ... [/MATH].

 

[blue_raver22]

Thank you!
Yes, you are on the right track. To show that A and B are measurable, we need to show that the outer measure is equal to the inner measure, i.e. \lambda^{*}(A) = \lambda_{*}(A) and \lambda^{*}(B) = \lambda_{*}(B).

To prove this, we can use the fact that for any set A, we have \lambda^{*}(A) = \inf\{\lambda(U) | U \supseteq A, U \text{ is open}\} and \lambda_{*}(A) = \sup\{\lambda(K) | K \subseteq A, K \text{ is compact}\}.

Now, since A \cup B is measurable, we have \lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B). This means that for any \epsilon > 0, there exist open sets U_1 and U_2 such that A \subseteq U_1, B \subseteq U_2, and \lambda(U_1) + \lambda(U_2) < \lambda(A \cup B) + \epsilon.

Similarly, since A \cup B is measurable, we also have \lambda(A \cup B) = \lambda_{*}(A) + \lambda_{*}(B). This means that for any \epsilon > 0, there exist compact sets K_1 and K_2 such that K_1 \subseteq A, K_2 \subseteq B, and \lambda(K_1) + \lambda(K_2) > \lambda(A \cup B) - \epsilon.

Now, since \lambda(A \cup B) = \lambda^{*}(A) + \lambda^{*}(B) = \lambda_{*}(A) + \lambda_{*}(B), we can combine the above inequalities to get \lambda(U_1) + \lambda(U_2) < \lambda(K_1) + \lambda(K_2) + 2\epsilon.

Since this is true for any \epsilon > 0, we can let \epsilon \to 0 to get \lambda(U_1) + \lambda(U_2) \leq \lambda(K_1) + \lambda(K_2).

But since U_1 and U_2 are