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.

  • $\begingroup$ $|B \setminus A| = |B| - |A|$ isn't valid if $|B|$ and $|A|$ are infinite. $\endgroup$ – Bungo Oct 7 '18 at 4:52
  • $\begingroup$ 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 $|B|\leq |C\cup D|$ $\leq |C|+|D|$ $\leq (|A|+\epsilon /2)+\epsilon /2=$ $|A|+\epsilon.$ $\endgroup$ – DanielWainfleet Oct 7 '18 at 10:59
  • $\begingroup$ $\sup \{|A|: A=\bar A\subset B\}$ is called the inner measure of $B$. $\endgroup$ – DanielWainfleet Oct 7 '18 at 11:05
  • $\begingroup$ @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. $\endgroup$ – Thy Art is Math Oct 7 '18 at 23:58

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