$A\subset\mathbb{R}, t>0\Rightarrow |A|=|A\cap (-t,t)|+|A\cap (\mathbb{R}-(-t,t))|$ 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.
 A: 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 $I_n'=I\cap(-t,t)$, $I_n''=I\cap(-\infty,-t]$, and $I_n'''=I\cap[t,\infty)$. 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 $\epsilon\to0$.
A: I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 8 on p.23 in Exercises 2A in this book.
David C. Ullrich, thank you very much for your answer.
I slightly modified your answer.
If $|A|=\infty$, then it is obvious that $$|A|=|A\cap (-t,t)|+|A\cap (\mathbb{R}\setminus (-t,t))|$$ holds.
We assume that $|A|<\infty$.
Obviously, $A=(A\cap (-t,t))\cup (A\cap (\mathbb{R}\setminus (-t,t)))$ holds.
So, by 2.8 (countable subadditivity of outer measure) on p.17 in this book, $$|A|\leq |A\cap (-t,t)|+|A\cap (\mathbb{R}\setminus (-t,t))|.$$
Note that $(A\cap (\mathbb{R}\setminus (-t,t)))\setminus (A\cap (\mathbb{R}\setminus [-t,t]))$ is $\emptyset$ or $\{t\}$ or $\{-t\}$ or $\{-t,t\}$.
So, $|(A\cap (\mathbb{R}\setminus (-t,t)))\setminus (A\cap (\mathbb{R}\setminus [-t,t]))|=0$.
So, $|A\cap (\mathbb{R}\setminus (-t,t))|\leq|(A\cap (\mathbb{R}\setminus [-t,t]))|$ by 2.8 (countable subadditivity of outer measure) on p.17 in this book.
And $|A\cap (\mathbb{R}\setminus [-t,t])|\leq|(A\cap (\mathbb{R}\setminus (-t,t)))|$ by 2.5 (outer measure preserves order) on p.16 in this book.
Therefore $|A\cap (\mathbb{R}\setminus [-t,t])|=|A\cap (\mathbb{R}\setminus (-t,t))|$.
Let $\epsilon$ be an arbitrary positive real number.
Suppose $I_1,I_2,\dots$ is a sequence of open intervals such that $\sum_{k=1}^\infty \mathcal{l}(I_k)<|A|+\epsilon$ and $A\subset\bigcup_{k=1}^\infty I_k$.
Let $J_k:=I_k\cap (t,\infty)$.
Let $K_k:=I_k\cap (-\infty, -t)$.
Let $L_k:=I_k\cap (-t,t)$.
Then, obviously, $\mathcal{l}(I_k)=|I_k|=|I_k\cap [t,\infty)|+|I_k\cap (-\infty, -t]|+|I_k\cap (-t,t)|=|J_k|+|K_k|+|L_k|=\mathcal{l}(J_k)+\mathcal{l}(K_k)+\mathcal{l}(L_k)$ holds. (We used the result of Exercise 6 on p.23 in Exercises 2A.)
Obviously, $J_1,K_1,J_2,K_2,\dots$ is a sequence of open intervals such that $A\cap (\mathbb{R}\setminus [-t,t])\subset J_1\cup K_1\cup J_2\cup K_2\cup\dots$.
Obviously, $L_1,L_2,\dots$ is a sequence of open intervals such that $A\cap (-t,t)\subset L_1\cup L_2\cup\dots$.
So, $|A\cap (-t,t)|+|A\cap (\mathbb{R}\setminus (-t,t))|=|A\cap (-t,t)|+|A\cap (\mathbb{R}\setminus [-t,t])|\leq (\mathcal{l}(J_1)+\mathcal{l}(K_1)+\mathcal{l}(J_2)+\mathcal{l}(K_2)+\dots)+(\mathcal{l}(L_1)+\mathcal{l}(L_2)+\dots)=(\mathcal{l}(J_1)+\mathcal{l}(K_1)+\mathcal{l}(L_1))+(\mathcal{l}(J_2)+\mathcal{l}(K_2)+\mathcal{l}(L_2))+\dots=\mathcal{l}(I_1)+\mathcal{l}(I_2)+\dots<|A|+\epsilon.$
Since $\epsilon$ was an arbitrary positive real number, $|A\cap (-t,t)|+|A\cap (\mathbb{R}\setminus (-t,t))|\leq |A|.$
Therefore $$|A|=|A\cap (-t,t)|+|A\cap (\mathbb{R}\setminus (-t,t))|$$ holds.
