Egoroff's Theorem-related Let $f_n: \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue measurable for $n \geq 1$.
Suppose $f(x)=\lim_{n \rightarrow } f_n(x)$ exists almost everywhere.
Show that if $\epsilon>0$, then there exists a measurable set $E \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$ for every $n \geq 1$.
This is sort of like Egoroff's Theorem. But $m(\mathbb{R})=\infty$ in this case. So we can't use Egoroff's theorem.
I tried to follow the prove of Egoroff's Theorem. But not sure how to proceed, because the measure of $\mathbb{R}$ is not finite.
Thanks in advance!
 A: First let us clarify the notation in the question.
The question is:

Let $f_i: \mathbb{R} \rightarrow \mathbb{R}$ be Lebesgue measurable for $i \geq 1$.
Suppose $f(x)=\lim_{i \rightarrow \infty} f_i(x)$ exists almost everywhere.
Show that, for every $n \geq 1$, if $\epsilon>0$, then there exists a measurable set $E \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$.

Given any $n \geq 1$, we have that $m([-n,n]) <+\infty$. So we can apply Egoroff's Theorem to $f_i$ and $f$ restricted to $[-n,n]$.
By apply Egoroff's Theorem to to $f_i$ and $f$ restricted to $[-n,n]$, we have that:
if $\epsilon>0$, then there exists a measurable set $F \subset [-n,n]$ with $m([-n,n]-F)< \epsilon$ and the convergence is uniform on $F$.
Now, take $E= [-n,n]-F$. Then, since $F \subset [-n,n]$, we have $F= [-n,n]-E$.
So, we have that: if $\epsilon>0$, then there exists a measurable set $E \subset [-n,n] \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$.
So, we have proved that: for every $n \geq 1$, if $\epsilon>0$, then there exists a measurable set $E \subset \mathbb{R}$ with $m(E)< \epsilon$ so that the convergence is uniform on $[-n,n ] - E$.
