Prove that if $B\subset \mathbf{R}$ is Lebesgue measurable, then $|B|=\sup\{|A|: A \text{ is a closed bounded subset of$B$}\}$.

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.

• $|B \setminus A| = |B| - |A|$ isn't valid if $|B|$ and $|A|$ are infinite. – Bungo Oct 7 '18 at 4:52
• Given $\epsilon\in \Bbb R^+,$ take $A=\bar A\subset B$ with $|B\setminus A|<\epsilon /3.$ Take an open set $C\supset A$ with $|C|\leq |A|+\epsilon/2.$ Take an open set $D\supset B \setminus A$ with $|D|\leq \epsilon/2.$ Then $C \cup D$ is open and $C\cup D \supset B$ so $|B|\leq |C\cup D|$ $\leq |C|+|D|$ $\leq (|A|+\epsilon /2)+\epsilon /2=$ $|A|+\epsilon.$ – DanielWainfleet Oct 7 '18 at 10:59
• $\sup \{|A|: A=\bar A\subset B\}$ is called the inner measure of $B$. – DanielWainfleet Oct 7 '18 at 11:05
• @DanielWainfleet I appreciate the help, but I'm having a hard time "reading" your comment. I guess I see alot of epsilons and I'm not sure why most of them were allowed or necessary. – Thy Art is Math Oct 7 '18 at 23:58