What's the difference between $\lim_{n \to \infty} \mu ([n,+\infty))$ and $\mu (\lim_{n \to \infty}[n,+\infty))$?
Some of my friends say that $\lim_{n \to \infty} \mu ([n,+\infty))=+\infty$ and $\mu (\lim_{n \to \infty}[n,+\infty))=0$.
I don't know whether it's right. Is there any reference for this?
On
Given a sequence of sets $A_n$ with $A_{n+1}\subset A_n$, one defines $$ \lim_{n\to\infty}A_n=\bigcap_{n=1}^\infty A_n $$ When you have $A_n\subset A_{n+1}$, you use union instead of intersection.
In your case, if $A_n=[n,\infty)$, then $A_{n+1}\subset A_n$, so $$ \lim_{n\to\infty} A_n=\bigcap_n A_n=\emptyset $$ On the other hand, each $A_n$ has infinite (Lebesgue) measure.
I'm assuming $\mu$ denotes the Lebesgue measure on $\mathbb R$.
First, let's address $$\lim_{n \to \infty} \mu ([n,+\infty))=+\infty$$ Clearly, $\mu ([n,+\infty))=+\infty$ for all $n\in\mathbb Z$, so the limit has to be $+\infty$.
When it comes to $$\mu (\lim_{n \to \infty}[n,+\infty))=0\tag1$$ it depends on what $\lim_{n \to \infty}[n,+\infty)$ mean as a set. One way to think about it is to consider the lim sup of sets: $$\limsup A_n= \bigcap_{N=1}^\infty \left( \bigcup_{n\ge N} A_n \right)$$ In that case, clearly, $$\lim\sup[n, +\infty)=\emptyset$$ which leads to $(1)$.
Another way to interpret $(1)$ is to consider the indicator function $\mathbb{1}_{[n;+\infty)}$ of $[n, +\infty)$. Then $(1)$ means $$\int_{\mathbb R}\left(\lim_{n\rightarrow+\infty} \mathbb{1}_{[n;+\infty)}\right)d\mu = 0$$ which is true by the dominated convergence theorem.