2
$\begingroup$

Let $E=[0,1]$. Here are the definitions I am using:

Let $A\subset E$, then we define the outer measure of $A$ as $$\mu^*(A)=\inf \left\{\sum_k m(I_k): A\subset \cup_k I_k \right\}$$ where the infimum is taken over all any countable collection $\{I_k\}$ of intervals (open, closed or half open) whose union contains $A$, and we define the inner measure of $A$ as, $$\mu_*(A)=1-\mu^*(E\setminus A)$$ and finally $A$ is said to be measurable if $\mu^*(A)=\mu_*(A)$.

As a note, I have shown that $$\mu^*(A)=\inf\{\mu(G): A\subset G, G \text{ is open relative to } E\}$$ so that we may use this characterization of the outer measure of a set or the one originally given above. We can also reformulate the definition of a measurable set: a set $A\subset E$ is measurable if and only if $\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A)$ and this follows immediately from the fact that if $A$ is measurable according to the definition above, then $\mu^*(A)+\mu^*(E\setminus A)=1$ and since $A\subset E$, we have $A\cap E=A$.

Now, I want to show yet another equivalent characterization of measurability in the following,

$\textbf{Problem:}$ I am trying to prove if $A \subset E$ is measurable then for any $F\subset E$, we have $$\mu^*(F)=\mu^*(F\cap A)+\mu^*(F\setminus A).$$

The hint in my book says to use $B\subset E$ is measurable if and only if for any $\epsilon >0$, $\exists G_1, G_2 \subset E$ and open relative to $E$ such that $B\subset G_1$, $E\setminus B\subset G_2$, and $\mu(G_1\cap G_2)<\epsilon$, so I also proved this (using the second characterization of $\mu^*$ provided here). For the current problem, this is my work so far:

Clearly, $$\mu^*(F)\leq \mu^*(F\cap A)+\mu^*(F\setminus A)$$ by sub-additivity. For the other inequality, let $\epsilon >0$, then by the above condition for measurability, we have (for some $G_1, G_2 \subset E$, etc), $$F\cap A\subset A \subset G_1 \text{ and } F\setminus A\subset E\setminus A\subset G_2$$ and $$\mu^*(F)+\mu^*(F\setminus A)\leq \mu(G_1)+\mu(G_2)=\mu(G_1 \cup G_2) + \mu(G_1\cap G_2)$$ but $G_1\cup G_2=E$, so the right hand side above is less than $1+\epsilon$. But I don't really see how this helps. We can conclude the left hand side is less than or equal to $1$ then since $\epsilon$ was arbitrary but I don't think I'm approaching this correctly. Any suggestions would be greatly appreciated, thanks.

$\endgroup$
  • $\begingroup$ You are trying to prove what is usually the definition of measurability. So it seems you are using a different (probably equivalent) definition of of measurability. What is the definition you are using? $\endgroup$ – Ramiro Dec 7 '15 at 14:27
  • $\begingroup$ @Ramiro I am sorry, edited to reflect which definition of measure I was using. For convenience, my def is $A\subset E$ is measurable if it satisfies $\mu_*(A)=\mu^*(A)$ where the inner measure is $1-\mu^*(E\setminus A)$ $\endgroup$ – Nap D. Lover Dec 7 '15 at 14:48
  • $\begingroup$ You wrote: "We can also reformulate the definition of a measurable set: a set $A\subset E$ is measurable if and only if $\mu^*(A)=\mu^*(E\cap A)+\mu^*(E\setminus A)$ and this follows... ". I think you meant $\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A)$. $\endgroup$ – Ramiro Dec 7 '15 at 21:18
  • $\begingroup$ @Ramiro yes, thanks for spotting the typo! $\endgroup$ – Nap D. Lover Dec 7 '15 at 21:26
1
$\begingroup$

Let $E=[0,1]$.

You already know that that $A\subset E$ is measurable (that is: $\mu^*(A)=\mu_*(A)$) if and only if
$$ \mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A)$$

You also know that:

  1. if $A\subset E$ then $$\mu^*(A)=\inf\{\mu(G): A\subset G, G \text{ is open relative to } E\}$$

  2. $A\subset E$ is measurable if and only if for any $\epsilon >0$, $\exists G_1, G_2 \subset E$ and open relative to $E$ such that $A\subset G_1$, $E\setminus A\subset G_2$, and $\mu(G_1\cap G_2)<\epsilon$ (The hint from your book is to use this property).

  3. $\mu^*$ is subadditive.

In the proof below we will use properties 1, 2 and 3 and also the following four properties:

  1. $\mu^*$ is monotone (that means, if $A\subset B$ then $\mu^*(A)\leqslant \mu^*(B)$).

  2. if $G \subset E$ is open relative to $E$, then $\mu^*(G)=\mu(G)$.

  3. if $G,H \subset E$ is open relative to $E$, then $\mu^*(G\setminus H)=\mu(G\setminus H)$.

  4. $\mu$ is additive (in fact, it is $\sigma$-additive).

Proof: Let $A\subset E$ be measurable and let $F\subset E$. Let $\epsilon >0$.

Since $\mu^*(F)<\infty$, then, by property 1, there is $G_0$ open set relative to $E$, such that $F\subset G_0$ and $\mu^*(F)\leqslant\mu(G_0)<\mu^*(F)+\epsilon$.

Since $A$ is measurable, then, by property 2, $\exists G_1, G_2 \subset E$ and open relative to $E$ such that $A\subset G_1$, $E\setminus A\subset G_2$, and $\mu(G_1\cap G_2)<\epsilon$.

Let $D=E\setminus G_2$. Then $D\subset A$ and we have \begin{align} \mu^*(F\cap A)&+\mu^*(F\setminus A)\leqslant \mu^*(G_0\cap A)+\mu^*(G_0\setminus A) \leqslant & \textrm{ prop. 4} \\& \leqslant \mu^*(G_0\cap G_1)+\mu^*(G_0\setminus D) \leqslant & \textrm{ prop. 4} \\& \leqslant \mu^*(G_0\cap D)+\mu^*(G_0\cap (G_1\setminus D))+\mu^*(G_0\setminus D) = & \textrm{ prop. 3} \\& = \mu^*(G_0\setminus G_2)+\mu^*(G_0\cap (G_1\cap G_2))+\mu^*(G_0\cap G_2) = & \textrm{ definition of $D$} \\& = \mu(G_0\setminus G_2)+\mu(G_0\cap (G_1\cap G_2))+\mu(G_0\cap G_2) = & \textrm{ prop. 5 and 6} \\& = \mu(G_0)+\mu(G_0\cap (G_1\cap G_2)) \leqslant & \textrm{ prop. 7} \\& \leqslant \mu(G_0)+\mu(G_1\cap G_2) \leqslant & \textrm{ prop. 4} \\& \leqslant \mu^*(F)+\epsilon + \epsilon \end{align}

So, for any arbritary $\epsilon>0$, we have $\mu^*(F\cap A)+\mu^*(F\setminus A) \leqslant \mu^*(F)+2\epsilon$. So we have $$ \mu^*(F)\geqslant \mu^*(F\cap A)+\mu^*(F\setminus A)$$ Since $\mu^*$ is subadditive, we get $\mu^*(F)= \mu^*(F\cap A)+\mu^*(F\setminus A)$.

$\endgroup$
  • $\begingroup$ Thank you! Very clear answer, much appreciated. $\endgroup$ – Nap D. Lover Dec 8 '15 at 17:54
  • $\begingroup$ I am sorry, I actually don't understand the step which uses prop 3 ($\sigma$-sub additivity), i am trying to figure it out, there must be some set relation i'm missing, could you elaborate? $\endgroup$ – Nap D. Lover Dec 9 '15 at 1:25
  • $\begingroup$ Sure. Note that $$G_1=D+(G_1\setminus D)$$ (where "$+$" indicates disjoint union). So $$G_0\cap G_1=(G_0\cap D)+(G_0\cap (G_1\setminus D))$$ Then, since $\mu^*$ is subadditive, we have $$\mu^*(G_0\cap G_1) \leqslant \mu^*(G_0\cap D)+\mu^*(G_0\cap (G_1\setminus D))$$ Please, let me know if you have any further question. $\endgroup$ – Ramiro Dec 9 '15 at 1:53
  • $\begingroup$ ah okay. Thanks again. Yet another question: I had not proven subadditivity for the formal disjoint union operation $+$ (or seen it used in my texts on measure) is there any difference than proving it for regular unions? $\endgroup$ – Nap D. Lover Dec 12 '15 at 23:28
  • $\begingroup$ For the sake of outer measures, no. It is just an immediate consequence of subadditivity for "regular" unions. $\endgroup$ – Ramiro Dec 12 '15 at 23:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.