Proving outer measure property

Question

I am self-studying analysis by Sheldon Axler. This is the one of exercise problem in his book. He uses $|\cdot|$ to indicate the outer measure.

Prove that if $A\subset \mathbb{R}$ and $t>0$, then $|A|=|A\cap(-t, t)|+|A\cap(\mathbb{R}\setminus(-t, t))|$.

$|A|\leq|A\cap(-t, t)|+|A\cap(\mathbb{R}\setminus(-t, t))|$ is obvious. But how do I prove inequality from the opposite side?

And in his next exercise, he somehow extends the property:

Prove that $|A|=\lim_{t\rightarrow\infty}|A\cap(-t, t)|$ for all $A\subset\mathbb{R}$.

Does this problem related to the previous problem? Any hints would be appreciated. Thanks

Answer 1

For the first part note that exists a measurable set $G \supset A$ which is $G_{\delta}$ such that $|A|=|G|$

Indeed by the definition of the outer measure(with coverings of open intervals) we can find open sets $G_n \supset A$ such that $|G_n| \leq |A|+\frac{1}{n}$

Sending $n \to +\infty$ we have that $|G|=|A|$

Take $G=\bigcap_nG_n$ and you have that $G \supset A$ and $|G| \leq |G_n| \leq |A|+\frac{1}{n}$

Then use the subadditivity of the outer measure and the measurability of $G$

Answer 2

Hint: by definition of outer measure, there is a sequence of intervals $I_n$ such that $A\subseteq \bigcup_n I_n$ and $|A|>\sum_n|I_n|-\epsilon.$ Consider $\{I_n\cap (-t,t)\}_n$ and $\{\mathbb R\setminus I_n\cap (-t,t)\}_n$