I am currently working on the following exercise
Prove that if $B\subset \mathbf{R}$ is Lebesgue measurable, then $$|B|=\sup\{|A|: A \text{ is a closed bounded subset of $B$}\}.$$
Here is my attempt at the proof so far:
Proof. If $B\subset \mathbf{R}$ is Lebesgue measurable, then for each $\epsilon>0$, there exists a closed set $A\subset B$ such that $|B\setminus A|<\epsilon$. Since outer measure is a measure on $(\mathbf{R},\mathcal{L})$, we may write $|B\setminus A|=|B|-|A|$ and thus the statement $|B\setminus A|<\epsilon$ can now be expressed as $|B|-|A|<\epsilon$. Rearranging this statement further yields $$|B|-\epsilon<|A|.$$ Using this with the fact that $|A|\leq |B|$ (because $A\subset B$), it follows that $$|B|=\sup\{|A|: A \text{ is a closed subset of } B\}.$$
My first question is: does this proof seem correct so far? Secondly, if this is indeed the correct approach, how in the world do I establish that $A$ is bounded?
Edit: The notation $\mathcal{L}$ stands for the set of all Lebesgue measurable sets and the notation $|B|$ and $|A|$ means the outer measure of $B$ and $A$, respectively.