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.)

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