## 515 – Caratheodory’s characterization of measurability (Homework 3)

This set is due Friday, April 27.

The goal of these problems is to prove Carathéodory‘s theorem that “extracts” a measure from any outer measure. In particular, when applied to Lebesgue outer measure, this construction recovers Lebesgue measure.

Recall that an outer measure on a set $X$ is a function $\lambda^*:{\mathcal P}(X)\to[0,+\infty]$ such that:

1. $\lambda^*(\emptyset)=0$.
2. $A\subseteq B\subseteq X$ implies $\lambda^*(A)\le\lambda^*(B)$.
3. For any $A_1,A_2,\dots$ subsets of $X$, we have $\lambda^*(\bigcup_n A_n)\le\sum_n\lambda^*(A_n)$.

Given a set $X$ and an outer measure $\lambda^*$ on $X$, let $\Sigma$ denote the collection of subsets $E$ of $X$ with the property that

$\lambda^*(A)=\lambda^*(A\cap E)+\lambda^*(A\setminus E)$

for all $A\subseteq X$.

• Prove that $\Sigma$ is a $\sigma$-algebra on $X$.

This requires some work. You may want to proceed by stages:

1. First, check that $\Sigma$ is precisely the collection of sets $E$ such that, for any $A\subseteq X$, we have

$\lambda^*(A)\ge\lambda^*(A\cap E)+\lambda^*(A\setminus E)$.

2. Check that $\emptyset \in\Sigma$, and that $\Sigma$ is closed under complements.
3. Check that $\Sigma$ is closed under finite unions. Conclude that it is also closed under set theoretic differences: If $E,F\in\Sigma$, then $E\setminus F\in \Sigma$.
4. The crux of the matter, of course, is to verify that $\Sigma$ is closed under countable unions. Accordingly, suppose that $E_n\in\Sigma$ for all $n$, and let $A\subseteq X$.

Let $E=\bigcup_n E_n$, and note that $E=\bigcup_n F_n$, where $G_n=\bigcup_{m\le n}E_m$, $F_1=G_1=E_1$ and, recursively, $F_n=G_n\setminus G_{n-1}=E_n\setminus G_{n-1}$ for $n>1$. (Note also that $G_n,F_n\in \Sigma$ for all $n$.)

Then, for $n>1$, $\lambda^*(A\cap G_n)=\lambda^*(A\cap G_{n-1})+\lambda^*(A\cap F_n)$, and for all $n$,

$\lambda^*(A\cap G_n)=\sum_{m\le n}\lambda^*(A\cap F_m).$

Conclude that $\lambda^*(A\cap E)\le\lim_{n\to\infty}\lambda^*(A\cap G_n)$. (Why does this limit exist?)

Also, prove that $\lambda^*(A\setminus E)\le\lim_{n\to\infty}\lambda^*(A\setminus G_n)$. (Again, why does this limit exist?)

Conclude from these inequalities and item 1 that $E\in\Sigma$. This concludes the proof that $\Sigma$ is a $\sigma$-algebra.

Now let $\lambda:\Sigma\to[0,+\infty]$ denote the restriction of $\lambda^*$ to $\Sigma$.

• Prove that $(X,\Sigma,\lambda)$ is a measure space.

In view of what we have proved already, note that this “reduces” to prove that, whenever $E_1,E_2,\dots$ are pairwise disjoint elements of $\Sigma$, then

$\lambda(\bigcup_n E_n)\ge\sum_n\lambda(E_n)$.

With notation as before, check first that $\lambda(G_n)=\sum_{m\le n}\lambda(E_n)$ for all $n$, and conclude.

• Prove that $\lambda$ is in fact a complete measure. Recall that this means that any subset of a set of $\lambda$-measure 0 is measurable and also has measure 0. In fact, check that if $\lambda^*(A)=0$, then $A\in\Sigma$, and conclude from this.
• Suppose that $Y\subseteq X$. Show that the restriction of $\lambda^*$ to ${\mathcal P}(Y)$ is an outer measure on $Y$. Denote by $\Sigma_X,$ resp. $\Sigma_Y$ the set $\Sigma$ defined above, for $\lambda^*,$ resp. $\lambda^*\upharpoonright{\mathcal P}(Y)$. Show that if $E\in\Sigma_X$, then $E\cap Y\in\Sigma_Y$. Suppose that $Y$ is measurable (i.e., that $Y\in\Sigma_X$). Is $\Sigma_Y=\Sigma_X\cap{\mathcal P}(Y)$? If so, is this the only case where equality holds?
• Prove that if $\lambda^*$ is $m^*$, Lebesgue outer measure on ${\mathbb R}^n$, then $\lambda$ is precisely $m$, Lebesgue measure on ${\mathbb R}^n$. (This may be a bit easier for $n=1$ than in general.)