Proving the *Caratheodory Criterion* for *Lebesgue Measurability*

Question

I'm following Terry Tao's An introduction to measure theory. Here he has defined lebesgue measurability as:

Definition 1 (Lebesgue measurability): A set $E$ is said to be lebesgue measurable if for every $\epsilon > 0$, $\exists$ an open set $U$, containing $E$, such that $m^*(U \setminus E) \leq \epsilon$.

where $m^*$ denotes the outer lebesgue measure.

Then he has stated the Caratheodory Criterion:

Definition 2 (Lebesgue measurability): A set $E$ is said to be lebesgue measurable if for every elementary set $A$, we have $$m(A)=m^*(A \cap E)+m^*(A \setminus E)$$

where an elementary set is a finite union of boxes.

The problem is to show the equivalence of the two definitions. I have shown that Def $(1)$ $\implies$ Def $(2)$, but having difficulty in showing the reverse, i.e. Def $(2)$ $\implies$ Def $(1)$. Any help (full/brief solution or even some hints) would be greatly appreciated! Thanks in advance.

Answer 1

Note: I just realized that your statement of Carathéodory's criterion doesn't agree with the usual one, where one tests against any possible test $A$. If we show that it is possible to test using open sets of finite measure (see @tomasz' answer), then we can do the following:

1. Lemma: for any set $E$ and any $\epsilon > 0$, there exists some open set $U \supset E$ such that $m^{\ast} (U) \le m^{\ast} (E) + \epsilon$. To prove this, note first that we can assume $m^{\ast} (E) < \infty$ (otherwise, choose U to be the whole space). By the definition of the exterior measure, we can find a countable family of (bounded) boxes $B_n, n \in \mathbb{N}$ covering $E$ and such that $\sum_n |B_n | < m^{\ast} (E) + \epsilon / 2$. Now enlarge each of them by $\epsilon / 2^{n + 1}$, creating open sets $U_n$ (e.g. if we are in $\mathbb{R}$, the boxes are of the form $B_n = [a_n, b_n)$ so we set $U_n = (a_n - \epsilon / 2^{n + 1}, b_n)$). Setting $U = \bigcup_n U_n$ we have by the subadditivity of $m^{\ast}$:

$$ m^{\ast} (U) \leq \sum_{n \in \mathbb{N}} m^{\ast} (U_n) = \sum_{n \in \mathbb{N}} | B_n | + \epsilon / 2 < m^{\ast} (E) + \epsilon. $$

2. To prove Def 2 $\rightarrow$ Def 1, fix $\epsilon > 0$, let $E$ fulfill Carathéodory's condition, and assume first $m (E) < \infty$. Consider the open set $U \supset E$ from step 1: we have $m^{\ast} (U) \leqslant m^{\ast} (E) + \epsilon$ and by assumption it holds that $m^{\ast} (U) = m^{\ast} (U \cap E) + m^{\ast} (U \backslash E)$, hence $m^{\ast} (U \backslash E) = m^{\ast} (U) - m^{\ast} (E) < \epsilon$. Finally, if $m(E) = \infty$, we cover it by a countable union of disjoint boxes $B_n$ of volume $1$ (e.g. the unit intervals in $\mathbb{R}$). We obtain disjoint sets $E_n = B_n \cap E$ with $m^{\ast} (E_n) \leqslant 1$ by monotonicity. For every $n \in \mathbb{N}$ we can apply the finite case with $\epsilon / 2^n$ and find open sets $U_n \supset E_n$ such that $m^{\ast} (U_n \backslash E_n) < \epsilon / 2^n$. Let $U = \bigcup U_n$. Then clearly $E \subset U$ and

$$ m^{\ast} (U \backslash E) \leqslant \sum m^{\ast} (U_n \backslash E) \leqslant \sum m^{\ast} (U_n \backslash E_n) < \epsilon . $$

Answer 2

Take any $E$ satisfying the Carathéodory condition. Consider first the case where $m^*(E)$ is finite. The general case can be deduced by splitting $E$ into countably many pieces of finite outer measure.

1. Note that any open set of finite measure can be approximated by an elementary set, that is, if $U$ is open of finite measure we can find for any $\varepsilon>0$ an elementary set $A$ with $m(U\triangle A)<\varepsilon$. (This is an easy consequence of countable additivity and the fact that any open set is a countable union of intervals.) Deduce that the Carathéodory condition also works for open sets of finite measure in place of elementary sets.
2. Choose any $\varepsilon>0$ and an open set $U$ such that $U\supseteq E$ and $m(U)<m^*(E)+\varepsilon$ (you have this more or less by definition of outer measure). By Carathéodory, we have $$ m^*(E\cap U)+m^*(U\setminus E)=m(U),$$ but $E\cap U=E$, so in fact $$ m^*(E)+m^*(U\setminus E)=m(U)<m^*(E)+\varepsilon,$$ and, since $m^*(E)$ is finite, we obtain $$ m^*(U\setminus E)<\varepsilon.$$

Answer 3

I know that this is an old post but the existing answers don't seem to use the 'restricted' version of Caratheodory.

Pick an open set such that $m^*(O)\leq m^*(E) + \epsilon$ for some $\epsilon > 0$. $$m^*(O)=\color{red}{ \Sigma m^*(B_n) \geq \Sigma(m^*(B_n \cap E) + m^*(B_n-E))}\geq m^*(\cup B_n\cap E)+m^*(\cup B_n-E)\\=m^*(E)+m^*(O-E)$$ This leads to $$m^*(E)+\epsilon \geq m^*(O)=m^*(E)+m^*(O-E)$$ Or $$\epsilon \geq m^*(O-E)$$ We have not even used the Caratheodory criterion for elementary sets. We have only used them for boxes. We have used the fact that open sets are just disjoint, countable boxes and the countable subadditivity of outer measure. The place where Caratheodory criterion is used is highlighted in red.