I am a graduate student of Mathematics.In our measure theory course I encountered the definition of Lebesgue measurable set which is as follows:
$(1)$ A set $A\subset \mathbb R$ is Lebesgue measurable if $\exists$a closed set $ B\subset A$ such that $|A-B|=0$.
I want to know whether the following is equivalent to the above definition:
$(2)$ A set $A$ is Lebesgue measurable if for each $\epsilon>0$ there exists a closed set $B_\epsilon\subset A$ such that $|A-B_\epsilon|<\epsilon$.
Clearly $(1)$ implies $(2)$ but is the other direction true?
Best Answer
The first definition is (strange and ) not equivalent to the second one (which agrees with the usual definition). Let $A=\mathbb R \setminus \{0\}$ . Certainly $A$ is Lebesgue measurable according any standard definition and it does satisfy the property in (2). But there is no closed set $B \subset A$ with $|A-B|=0$. This is because $|B^{c}|=0$ (: $B^{c} \subset (A-B) \cup \{0\}$) which implies that $B$ is dense. But $B$ is closed, so $B=\mathbb R$. This contradicts the fact that $B \subset A$.