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.


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 $m^{\ast}$:

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

  1. 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

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

  • $\begingroup$ Can you show me a bit more? I tried but cannot find the right line of attack based on your hint. $\endgroup$ – Roronoa Dec 20 '16 at 2:40
  • $\begingroup$ Sure, I'm editing the question right now. $\endgroup$ – Miguel Dec 20 '16 at 9:57
  • $\begingroup$ This is beautiful :) Thanks a lot!! $\endgroup$ – Roronoa Dec 20 '16 at 14:41
  • $\begingroup$ I've one more question. The exercise considered a restricted version of the Caratheodory Criterion (In the book it is mentioned as Caratheodory Criterion: One Direction). If we employ a more stronger version of Caratheodory which holds for open set $U$ containing $E$ (instead of only elementary sets containing $E$), Def $(2)$ $\implies$ Def $(1)$ becomes easy. Is it possible to accomplish the goal using only the weaker version mentioned in the problem? Alternatively, is it possible to prove the stronger version used in the solution, using the weaker version only? $\endgroup$ – Roronoa Dec 21 '16 at 5:39

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 $$ m^*(U\setminus E)<\varepsilon.$$
  • $\begingroup$ Hmm... I agree that multiple answers are generally good but, if I may say so: I think that it would make more sense to prove your assertion in #1 that the Carathédory condition holds for open sets of finite measure and use it to complete my answer, which essentially does the same things, but in more detail (as requested) $\endgroup$ – Miguel Dec 20 '16 at 11:14

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