$A\subset\mathbb{R}, t>0\Rightarrow |A|=|A\cap (-t,t)|+|A\cap (\mathbb{R}-(-t,t))|$

Question

I am trying to prove the following statement from Axler's MIRA book and I would appreciate an hint about how to finish my proof (NOTE: $||$ refers to outer measure):

"$A\subset\mathbb{R}, t>0\Rightarrow |A|=|A\cap (-t,t)|+|A\cap (\mathbb{R}-(-t,t))|$"

What I have done up to now:

Let $t>0$ and $A$ be a subset of $\mathbb{R}$: then we have that

$$|A|=|A\cap\mathbb{R}|=|A\cap ((-,t,t)\cup\mathbb{R}-(-t,t))|=|(A\cap (-t,t))\cup (A\cap\mathbb{R}-(-t,t))|\leq |(A\cap (-t,t))|+|(A\cap\mathbb{R}-(-t,t))|$$ where the first equality follows from the fact that $A\subset\mathbb{R}$, the second from the fact that $\mathbb{R}=(-t,t)\cup (\mathbb{R}-(-t,t))$ and the inequality from the countable subadditivity of outer measure.

So, what I have to do now is find a way to get an inequality of the form $$|(A\cap (-t,t))|+|(A\cap\mathbb{R}-(-t,t))|\leq |A|$$ but I don't see how to do this, so I would appreciate an hint (not a solution) about how to do this.

Answer 1

It really is immediate from the definition. First a technicality:

Lemma The definition of $|A|$ is unchanged if closed or half-open intervals are allowed as well as open intervals.

Proof: Say $|A|'$ is the outer measure including all sorts of intervals. The inf of a larger set is smaller, so $|A|'\le|A|$.

To show the other inequality: If $I$ is any interval with endpoints $a<b$ let $B(I,\epsilon)=(a-\epsilon/2, b+\epsilon/2)$. Let $\epsilon>0$. Choose intervals $I_1,\dots$ so$$A\subset\bigcup I_n$$ and $$\sum l(I_n)<|A|'+\epsilon.$$Let $I_n'=B(I_n,\epsilon/2^n)$. Since $I_n'$ is open and $A\subset\bigcup I_n'$ we have $$|A|\le\sum l(I_n')=\sum(l(I_n)+\epsilon/2^n)=\epsilon+\sum l(I_n)\le|A|'+2\epsilon.$$ Since $\epsilon>0$ is arbitrary this shows $|A|\le|A|'$.

Now to show $|A\cap(-t,t)|+|A\setminus(-t,t)|\le|A|$: Wlog $|A|<\infty$. Let $\epsilon>0$. Choose intervals $I_1,\dots$ so $$A\subset\bigcup I_n$$ and $$\sum l(I_n)<|A|+\epsilon.$$ Let $I_n'=I_n\cap(-t,t)$, $I_n''=I_n\cap(-\infty,-t]$, and $I_n'''=I_n\cap[t,\infty)$. Then $A\cap(-t,t)\subset\bigcup I_n'$ and $A\setminus(-t,t)\subset\bigcup(I_n''\cup I_n'''), $ so $$|A\cap(-t,t)|+|A\setminus(-t,t)|\le\sum l(I_n')+\sum(l(I_n'')+l(I_n'''))=\sum l(I_n)\le|A|+\epsilon;$$as before, let $\epsilon\to0$.