about application of egoroff’ theorem $f_n:\mathbb{R}\rightarrow \mathbb{R}$ is measureable function ,for any x$\in \mathbb{R},\lim_{n\rightarrow \infty}f_n(x)=f(x)=1$,proof :there exist measureable set sequence {$E_k$},s.t. $f_n$ convergence uniformly to f ,and m($\mathbb{R}-\bigcup _kE_k$)=0
My attempt is cut R to some interval and use egoroff’s theorem but I don’t know how to do specific 
 A: There is nothing big difference with the last answer but I tried...
Let $A_n=[n,n+1]\cup [-n-1,-n]$, then $m(A_n)<\infty$. Then, by Egoroff's theorem, for each non-negative integer $n$,  there is a measurable set $E_{n,k}\subset A_n$ such that $m(A_n\setminus E_{n,k})<\frac{1}{2^n k}$ and $f_n\rightarrow f$ uniformly on $E_{n,k}$.  Then observe that, since $\mathbb{R}=\bigcup_{n=0}^\infty A_n$, for given $j\in \mathbb{Z}^+$,
\begin{align*}
m(\left(\bigcup_{n,k}E_{n,k}  \right)^c)&= m\left( \bigcup_{n} A_n \setminus \bigcup_{n,k} E_n  \right)\\ &= m\left(\bigcup_{n}\left(A_n\setminus \bigcup_{k} E_{n,k}\right)\right)\\&\leq \sum_n m\left(A_n\setminus \bigcup_k  E_n\right)\\&=\sum_n m\left(\bigcap_k\left( A_n\setminus E_{n,k} \right)  \right)\\& \leq \sum_n m(A_n\setminus E_{n,j})  \\&<\sum_n\frac{1}{2^nj}\\&<\frac{2}{j}.
\end{align*}
Then since $j\in \mathbb{Z}^+$ is arbitrary, $m(\left(\bigcup_{n,k}E_{n,k}  \right)^c)=0$. And $\{E_{n,k}\}$ is the sequence of measurable sets we were finding.
TMI(or called remark) : 
If, in addition, $f_n$ were continuous and $f_n\leq f_{n+1}$, by Dini's theorem, we could just say $f_n\rightarrow f$ unifomly on $A_n$ since $A_n$ is compact. Thus, in this case $\{A_n\}$ is the sequence of measurable sets we were finding.
