# Lebesgue measurability: Rudin's vs Tao's

I'm a bit confused about Lebesgue measurability of a set. I saw two different forms of definition:

(1) In Terence Tao's Introduction to Measure Theory, the definition is quite simple. First, the Lebesgue outer measure of any $E \subset \mathbb R^p$ is $m^*(E)\triangleq \underset{\{B_n\}}\inf \: \sum_{n=1}^\infty m(B_n)$, where $\cup B_n$ is a cover of $E$, consisting of boxes $B_n$, and $m$ is Jordan measure. Then, a subset $B$ of $\mathbb R^p$ is said to be Lebesgue measurable if for every $\epsilon > 0$, there exists an open set $U \subset \mathbb R^p$ such that $m^*(U\setminus B)<\epsilon.$

(2) In Walter Rudin's Principles of Mathematical Analysis, the definition is a bit more lengthy:
First, the Lebesgue outer measure is defined almost identically to that in Tao's, with the only difference being that $B_n$'s are open elementary sets.

Then, $A \subset \mathbb R^p$ is said to be $m$-finitely measurable, if there exists a sequence $\{A_n\}$ of elementary sets such that $A_n \to A,$ where $d(A, A_n) \triangleq m^*((A\setminus A_n)\cup (A_n\setminus A))$.

Finially, $B \subset \mathbb R^p$ is said to be Lebesgue measurable, if $B$ is a countable union of $m$-finitely measurable sets.

My questions are:

(1) Are these two definitions of Lebesgue measurability equivalent?
(2) If so, how may we prove it? (If the proof is long, I'd also greatly appreciate a reference.)
(3) It appears to me that Tao's definition is easier to use and comprehend, and also seems to be more prevalent nowadays. Is this true? Is Rudin's definition still useful in some cases?

I'd appreciate any help, comments, partial answers, ... etc. Thanks a lot!

P.S. This question is related to Equivalent definitions of Lebesgue Measurability (Rudin and Royden), which is still open. I compare Rudin's and Tao's, provide more details, and have more questions.

• (1) In fact, yes. One can prove that if we define measurable sets via symmetric difference (no. 2 in your list), then the outer measure (no. 1) is the only non-negative countably-additive extension of $m$ to the set of all measurable subsets. (2) For instance, you can find the equivalence proof in V. Bogachev book on measure theory (vol.1). – hyperkahler May 23 '17 at 4:06
• @Arteom Thanks a lot! I appreciate the pointers! – syeh_106 May 23 '17 at 4:12