Prove that $ |A| = \lim_{t\rightarrow \infty}| A \cap (-t,t)|$ for all $A \subset \mathbb{R}$ Problem taken from the books sheldon Axler Measure , integration Real analysis
Prove that $ |A| = \lim_{t\rightarrow \infty}| A \cap (-t,t)|$ for all $A \subset \mathbb{R}$
My attempt : $\lim_{t\rightarrow \infty}| A \cap (-t,t)|=|A \cap (-\infty,\infty)|=\min| A|$
Im newly learning measure theory
 A: Take a sequence $t_n\nearrow \infty$ (i.e., $t_n>t_{n-1}$) and let $A_n:=A\cap(-t_n,t_n)$. By monotonicity, $\lim_{n\to\infty}|A_n|\le |A|$ ($\because A_n\subseteq A$ for each $n\ge 1$ and $|A_n|$ is nondecreasing). Thus, it remains to show that
$$
\lim_{n\to\infty}|A_n|\ge |A|.\tag{1}\label{1}
$$
Let $B_n:=A_n\setminus A_{n-1}$, where $A_0\equiv \emptyset$. Using Problem 2A.8 in Axler's book, we have
\begin{align}
|A_n|&=\left|\bigcup_{i=1}^{n} B_{i}\right|=\left|\bigcup_{i=1}^{n} B_{i}\cap(-t_{n-1},t_{n-1})\right|+\left|\bigcup_{i=1}^{n} B_{i}\cap (\mathbb{R}\setminus(-t_{n-1},t_{n-1}))\right| \\
&=\left|\bigcup_{i=1}^{n-1} B_{i}\right|+\left|B_n\right|=\cdots= \sum_{i=1}^n |B_i|.
\end{align}
Consequently, $\sum_{n\ge 1}|B_i|=\lim_{n\to\infty}|A_n|$. If the sum on the RHS is infinite, the inequality $\eqref{1}$ is trivially satisfied. Otherwise, since $|\cdot|$ is subadditive, for each $m\ge 1$,
$$
|A|\le |A_m|+\sum_{n> m}|B_i|,
$$
and the second term on the RHS converges to $0$ as $m\to\infty$.
A: *

*If $A$ is measurable then monotone convergence gives the desire result, just as in d.k.o.'s proof.


*If $A$ is not measurable and $|A|^*=\infty$, then $$\lim_{t\rightarrow\infty}|A\cap(-t,t)|^*\geq\lim_{t\rightarrow\infty}|(-t,t)|=\infty$$
where $|\cdot|^*$ is the outer measure induced by $|\cdot|$ and the collection of countable unions of intervals of the form $(a,b]$, $-\infty<a<b<\infty$ (Caratheodory's construction)


*If $|A|^*<\infty$, then there is a measurable set $B$ such that $A\subset B$ and $|B\cap E|=|A\cap E|^*$ for every measurable set $E$. (This is Caratheodory's construction of measurability). Then
$$\lim_{t\rightarrow\infty}|A\cap(-t,t)|^*=\lim_{t\rightarrow\infty}|B\cap(-t,t)|=|B|=|A|^*$$

Comment:
The limit $\lim_{t\rightarrow\infty}$ can be understood as taking a limit over any increasing sequence $t_n\nearrow\infty$ as $n\rightarrow\infty$. That is to make use of monotone convergence (or $\sigma$-continuity) of the measure $|\cdot|$.
A: Hint:
Prove that the limit is equal to $\lim_{n\to \infty}|A\cap(-n,n)|$. Now use the fact that $\big(A\cap (-n,n)\big)$ is an increasing sequence.
A: user140541, thank you very much for your answer.
I very slightly changed your proof and I wrote a proof for me.　　
Let $f(t):=|A\cap (-t,t)|$ for any $t\in (0,\infty)$.
Then, $f(t)\leq |(-t,t)|=2t$.
So, $f(t)\in\mathbb{R}$ for any $t\in (0,\infty)$.
If $s<t$, then $f(s)\leq f(t)$.
If $\{f(t):t\in (0,\infty)\}$ is not bounded above, then $\lim_{t\to\infty} f(t)=\infty$.
If $\{f(t):t\in (0,\infty)\}$ is bounded above, then $\lim_{t\to\infty} f(t)=\sup\{f(t):t\in (0,\infty)\}\leq |A|$.

Let $A_0:=\emptyset$.
Let $A_n:=A\cap (-n,n)$ for any $n\in\{1,2,\dots\}$.
Let $B_n:=A_n\setminus A_{n-1}$ for any $n\in\{1,2,\dots\}$.
Then, $A_n=\bigcup_{k=1}^{n} B_k$.
By Exercise 8 in Exercises 2A in the book, $$|A_n|=|A_n\cap (-(n-1),n-1)|+|A_n\cap (\mathbb{R}\setminus (-(n-1),n-1))|\\=|A_{n-1}|+|B_n|\\=|A_1|+|B_2|+\dots+|B_n|\\=|B_1|+|B_2|+\dots+|B_n|.$$

If $\lim_{n\to\infty} f(n) = \lim_{n\to\infty} |A_n|=\infty$, then $|A|=\infty$ since $|A_n|\leq |A|$.
In this case, $\{f(t):t\in (0,\infty)\}$ is not bounded above.
So, $\lim_{t\to\infty} f(t)=\infty=|A|$.

Suppose $\lim_{n\to\infty} f(n) = \lim_{n\to\infty} |A_n|\in\mathbb{R}$.
Obviously, $A=\bigcup_{k=1}^\infty B_k$.
By countable subadditivity of outer measure, $$|A|\leq\sum_{k=1}^\infty |B_k|=\lim_{n\to\infty} |A_n|.$$
Therefore $\lim_{n\to\infty} f(n) = \lim_{n\to\infty} |A_n|=|A|$.

Let $\epsilon$ be an arbitrary positive real number.
There exists $N\in\{1,2,\dots,\}$ such that $|A|-\epsilon<f(N)\leq |A|$.
If $N<x$, then $|A|-\epsilon<f(N)\leq f(x)\leq |A|$.
Therefore $\lim_{x\to\infty} f(x)=|A|$.
