# Almost uniform convergence implies a.e. pointwise convergence proof

I've just read a proof of the statement "On a finite measurable space, $(f_n)_{n \geq 1}$ and $f$ measurable and finite a.e. functions, if $(f_n)_{n \geq 1}$ converges almost uniformly to $f$, then it converges pointwise to $f$ a.e." I have some doubts with the proof given, so I'll write the proof first:

We start by picking a set $F_m$ for every positive integer $m$ so that $\mu(F_m)<\frac{1}{m}$, and so that $\{f_n\}$ converges uniformly to $f$ on $\{F_m\}^c$. We set

$\displaystyle F=\bigcap\limits_{m=1}^\infty F_m$

and conclude that $\mu(F)=0$, since $\mu(F)\leq\mu(F_m)<\frac{1}{m}$ for all $m$. If $x\in F^c$, then there must be some $m$ for which $x\in{F_m}^c$. Since $\{f_n\}$ converges to $f$ uniformly on ${F_m}^c$, we conclude that $\{f_n(x)\}$ converges to $f(x)$. Thus $\{f_n\}$ converges pointwise to $f$ except on the set $F$ of measure zero.

Why the statement "Since $\{f_n\}$ converges to $f$ uniformly on ${F_m}^c$, we conclude that $\{f_n(x)\}$ converges to $f(x)$" holds?, I mean, I am thinking convergence in terms of $\epsilon-n_0$ proof, so I am having some difficulty understanding convergence on $F$ in these terms. I would really appreciate some help understanding this, thanks in advance.

Your sequence converges uniformly on each $F_m^c$. Hence it converges pointwise on each $F_m^c$. Hence it converges pointwise on the union of the $F_m^c$. (This is a basic special property of pointwise convergence: if $f_n$ converges pointwise on each set in a family, then it converges pointwise on their union.) But the union of the $F_m^c$ has full measure.
In terms of a $\varepsilon,n_0$ proof, let $x \in F^c$, then choose $m$ such that $x \in F_m^c$. Since you have uniform convergence on $F_m^c$, given $\varepsilon > 0$ you get $n_0$ so that if $x \in F_m^c$ and $n \geq n_0$ then $|f_n(x)-f(x)|<\varepsilon$.
• Thanks for your answer! I still have some doubts: given $\epsilon$, in each $F_m^c$ there is $n(\epsilon,m)$ such that $|f_n(x)-f(x)|<\varepsilon$, I don't see how this implies there exists $N$ such that for all $n \geq N$, $|f_n(x)-f(x)|<\epsilon$ IN $F^c$ – user156441 Apr 16 '15 at 4:06
• I mean, I don't understand why "if $f_n$ converges pointwise on each set in a family, then it converges pointwise on their union" is valid. – user156441 Apr 16 '15 at 4:08
• @user156441 The $n_0$ does not work for all of $F^c$, it depends on $x$ through the choice of $m$. Alternately, an explanation without explicit quantifiers is to notice that "$f_n$ converges to $f$ pointwise on $A$" means "for each $x \in A$, $f_n(x)$ converges to $f(x)$ in the sense of real numbers". – Ian Apr 16 '15 at 12:21