showing $f=g$ a.e. given $f_n\to f$ uniformly and $f_n\to g$ in $L^p(\Bbb R)$ Let $f_n: \Bbb R \to \Bbb R$ be continuous functions that converge uniformly to a function $f$.
Suppose that $f_n\to g$ for $g\in L^p(\Bbb R)$ for some $p\ge 1$.
I need to prove that $f=g$ a.e.
So it's enough to show that $||f-g||_p =0$.
By triangle inequality we have $||f-g||_p \le ||f-f_n||_p +||f_n-g||$. 
so it's enough to show that $||f-f_n||_p \to 0$ (as ,by assumption $||f_n-g||\to 0)$. 
So I reduced the problem to the problem of showing $||f_n-f||_p \to 0$ , any ideas how to do that?
I know $f$ is continuous and also that I can bound $|f_n(x)-f(x)|^p$ uniformly, but I still have an integral of a constant over $\Bbb R$ which is $\infty$.  
Thanks for helping!
 A: I don't think that starting from the $L^p$ bound will be very helpful. This is because you have much stronger hypotheses (uniform convergence for $(f_n)_n$) but you immediately go into the realm of integrals, thus losing some of the leverage you started with (convergence implications usually go the other way). So let's try it the other way around: it suffices to see that for each $\varepsilon > 0$, the set
$$
E_\varepsilon = \{|f-g| > \varepsilon\}
$$
is of measure zero. Since $|f-g| \leq |f-f_n| + |f_n-g|$, then $|f-g| > \varepsilon$ implies either $|f-f_n| > \varepsilon/2$ or $|f_n-g| > \varepsilon/2$. This gives
$$
E_\varepsilon \subset \{|f-f_n| > \varepsilon/2\} \cup \{|f_n-g| > \varepsilon/2\}
$$
and taking measures, 
$$
|E_\varepsilon| \leq |\{|f-f_n| > \varepsilon/2\}| + |\{|f_n-g| > \varepsilon/2\}|.
$$
Now you can use that $f_n \to f$ uniformly and $f_n \to g$ in $L^p$ (think of Tchevyscheb's inequality).
A: Let $N\in \mathbb{N}$. Then, 
$$
0=\lim_{n}\int_{-N}^N|f_n-g|^p\stackrel{\text{uniform convergence}}{=}\int_{-N}^N|f-g|^p
$$
so $f=g$ almost everywhere on $[-N,N]$. 
Suppose now there is some positive measure set $A$ on which $f$ and $g$ differ. Then, by continuity of measure, 
$$
\mu(A)=\mu\left (\bigcup_{N\in \mathbb{N}}[-N,N]\cap A\right)=\lim_{N\to \infty}\mu([-N,N]\cap A)
$$
So, for some large $N'$, $\mu([-N',N']\cap A)>0$, a contradiction.
A: Continuity and uniform convergence don't really have much to do with this. Assume $f_n$ is a sequence of measurable functions on $\mathbb R$ that converges pointwise to $f$ a.e. If also $f_n\to g$ in $L^p,$ then $f=g$ a.e.
Proof: We can find $A\subset \mathbb R$ with $m(A^c)=0$ such that $f_n\to f$ pointwise on $A.$ Recall that $f_n\to g$ in $L^p$ implies $f_{n_k}\to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $B\subset \mathbb R$ with $m(B^c)=0$ such that $f_{n_k}\to g$ pointwise on $B.$ On the set $A\cap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $A\cap B.$ Since $m[(A\cap B)^c] = 0,$ we're done.
