[Math] 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

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