9
$\begingroup$

This comes from an exercise from Real Analysis by Folland.

Let $\mathcal{A}\subset P(X)$ be an algebra, $\mathcal{A}_\sigma$ the collection of countable unions of sets in $\mathcal{A}$, and $\mathcal{A}_{\sigma\delta}$ the collection of countable intersections of sets in $\mathcal{A}_\sigma$. Let $\mu_{0}$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induced outer measure.

a.) For any $E\subset X$ and $\epsilon > 0$ there exists $A\in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(A) \leq \mu^*(E) + \epsilon$.

b.) If $\mu^{*}(E) < \infty$, then $E$ is $\mu^{*}$-measurable if and only if there exists $B\in A_{\sigma\delta}$ with $E\subset B$ and $\mu^{*}(B\setminus E) = 0$

c.) If $\mu_0$ is $\sigma$-finite, the restriction $\mu^{*}(E) < \infty$ in (b) is superfluous

proof a.): By the definition of outermeasure we know that $$\mu^{*}(E) = \inf\left\{ \sum_{j=1}^{\infty} \mu_{0} ( A_{j} ) : A_{j} \in \mathcal{A}, E \subset \bigcup_{j=1}^{\infty} A_{j} \right \}$$ Let $A = \bigcup_{j=1}^{\infty} A_{j}$ as above. Then $A \in \mathcal{A}_{\sigma}$ and $E \subset A$. For each $j$ we can construct a sequence $\{ B_{j}^{k} \} _{k=1}^{\infty}\subset \mathcal{A}$ such that $A_j\subset \bigcup_{j,k=1}^{\infty}B_{j}^{k}$. It follows that since $$\mu^{*}(A_j) = \inf\left\{ \sum_{j=1}^{\infty} \mu_{0} (B_{j}^{k} ) : B_{j}^{k} \in \mathcal{A}, A_j \subset \bigcup_{j,k=1}^{\infty} B_{j}^{k} \right \}$$ We have that, $$\mu^{*}(A_{j}) \leq \mu_{0}(A_{j}) + \epsilon 2^{-j}, \forall j \ \ \text{and} \ \ \epsilon>0$$ Thus: $$\mu^{*}(A) \leq \sum_{j=1}^{\infty} \mu^{*} ( A_{j} ) \leq \sum_{j=1}^{\infty} ( \mu_{0}(A_{j}) + \epsilon 2^{-j}) = \mu^{*}(E) + \epsilon$$ Since $\epsilon$ is arbitrary we are done.

b.) Suppose $E$ is $\mu^{*}$-measurable. From part (a), we know that $\forall n\in\mathbb{N}$ there exists $B_n\in A_\sigma$ with $E\subset B_n$ and $\mu^{*}(B_n) - \mu^{*}(E) \leq 1/n$. Let, $$B = \bigcap_{n = 1}^{\infty}B_n\in A_\sigma$$ since E is $\mu^{*}$-measurable, we have $\mu^{*}(B_n) = \mu^{*}(B_n\cap E) + \mu^{*}(B_n\cap E^{c})$ hence, $$\mu^{*}(B\cap E^{c})\leq \mu^{*}(B_n\cap E^{c})= \mu^{*}(B_n) - \mu^{*}(E)\leq 1/n$$ for every $n\in\mathbb{N}$. Hence we have $\mu^{*}(B\setminus E) = 0$\ To show the converse, let's suppose $B\in A_{\sigma\delta}$ with $E\subset B$ and $\mu^{*}(B\setminus E) = 0$. From part (a), we know that $\forall n\in\mathbb{N}$ there exists $A_n\in A_\sigma$ with $(B\setminus E)\subset A_n$ and $\mu^{*}(A_n) - \mu^{*}(B\setminus E)\leq 1/n$. But, since $\mu^{*}(B\setminus E) = 0$ then, $\mu^{*}(A_n)\leq 1/n$. Let, $$A = \bigcap_{n=1}^{\infty}A_n$$ then A is $\mu^{*}$-measurable (since $A\in A_{\sigma\delta}$ and the set of all $\mu^{*}$-measurable sets is a $\sigma$-algebra) such that $(B\setminus E)\subset A$ and $\mu^{*}(A) = 0$.\ By Carathedors's theorem we know that the restriction of $\mu^{*}$ to $\mu^{*}$-measurable sets is a complete measure. From this, we know that $(B\setminus E)$ is $\mu^{*}$-measurable. Also, since $B\in A_{\sigma\delta}$ then $B$ is also $\mu^{*}$-measurable and we can express $E$ as $$E = (B^{c}\cup (B\cap E^{c}))^{c}$$ Thus $E$ is $\mu^{*}$-measurable.

c.) Let $\mu_0$ be $\sigma$-finite, then let $$X = \bigcup_{1}^{\infty}X_i$$ where $X_i\in M$ and $\mu(X_i) < \infty$ Now, suppose $E$ is $\mu^{*}$-measurable and $\mu^{*}(E) = \infty$, set $$E_n = (E\cap \bigcup_{1}^{n}X_i)$$ then $\mu^{*}(E_n) < \infty$ and $E = \bigcup_{1}^{\infty}E_n$. Let $\epsilon > 0$, from part (a) $\forall n\in\mathbb{N} \exists C_n\in A_\sigma$ such that $E_n\subset C_n$ and $$\mu^{*}(C_n\setminus E_n) = \mu^{*}(C_n) - \mu^{*}(E_n) \leq \epsilon/2^{n}$$ $$\mu^{*}(E_n) = \mu^{*}(E\cap \bigcup_{1}^{n}X_i) = \mu^{*}(E) \cap \mu^{*}(\bigcup_{1}^{n}X_i) = \infty$$ hence $\mu^{*}(C_n\setminus E_n) = 0$

I am not sure where to go from here, any hints or suggestions is greatly appreciated

$\endgroup$
  • $\begingroup$ Every time you have the $\sigma$-finiteness assumption you do the same thing, which is you consider a sequence of increasing sets of finite measure such that the union is the entire space. You'll need to apply result (b) in each of these sets. I am pretty sure you can take it from here :) $\endgroup$ – Giovanni Aug 31 '15 at 0:43
  • $\begingroup$ @Giovanni check out what I tried, do you know where I need to go from here? $\endgroup$ – Wolfy Aug 31 '15 at 0:59
  • $\begingroup$ I don't want the comment section to get too messy, I'll write down a solution and post it :) $\endgroup$ – Giovanni Aug 31 '15 at 1:32
7
$\begingroup$

The solution for $(c)$ is perfect up to $\mu^{*}(C_n\setminus E_n) = \mu^{*}(C_n) - \mu^{*}(E_n) \leq \epsilon/2^{n}$, so let's keep going from here.

It is easy to lose track of what we are doing with these kind of problems, so let's not forget that we are trying to approximate the set $E$ from outside, and that right now we only have an approximation for every $n$. The obvious thing to do is then to consider the union of the approximating sets: let $$C_{\epsilon} = \bigcup_{n = 1}^{\infty}C_n.$$

The $\epsilon$ in $C_{\epsilon}$ is there to stress the dependence upon $\epsilon$ of the $C_n$'s and hence of $C$. It is worth noticing that $C_{\epsilon} \in \mathcal{A}_{\sigma}$ being the countable union of countable unions of elements of $\mathcal{A}$.

For the next step to work you might want to state clearly in your proof that we can take $C_n \subset \cup_{i = 1}^nX_i$.

Intuitively, $C_{\epsilon}$ is a decent approximation of $E$. Let's make this precise: \begin{align} \mu^*(C_{\epsilon} \cap E^c) = & \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E^c)\Big) = \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E_n^c)\Big) \\ \le & \sum_{n = 1}^{\infty}\mu^*(C_n \setminus E_n) \le \epsilon. \end{align}

(good job considering the $2^{-n}$, that really came in handy!)

Let's go back to our intuition: $C_{\epsilon}$ is an $\epsilon$-good approximation of $E$, therefore we would like to send $\epsilon$ to $0$ to get the best possible approximation. Since we want to end up in $\mathcal{A}_{\sigma \delta}$ what we need to do is to "discretize" the sets $C_{\epsilon}$ considering instead the sets $C_{\frac{1}{n}}$. (the notation is unfortunate, these resemble too much the $C_n$ sets, hopefully this won't cause any confusion)

Then clearly $$C := \bigcap_{n = 1}^{\infty}C_{\frac{1}{n}} \in \mathcal{A}_{\sigma \delta}$$ and satisfies $\mu^*(C \setminus E) = 0$, proving the claim.

To show this last equality notice that $$\mu^*(C \setminus E) \le \mu^*(C_{\frac{1}{n}} \setminus E) \le \frac{1}{n} \to 0.$$

Let me know if there is anything that I need to clarify!

$\endgroup$

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.