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Let $f_n: Bbb R to Bbb R$ be continuous functions that converge uniformly to a function $f$.

Suppose that $f_nto g$ for $gin L^p(Bbb R)$ for some $pge 1$.

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!

Let $Nin mathbb{N}$. Then,
$$
0=lim_{n}int_{-N}^N|f_n-g|^pstackrel{text{uniform convergence}}{=}int_{-N}^N|f-g|^p
$$
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,
$$
mu(A)=muleft (bigcup_{Nin mathbb{N}}[-N,N]cap Aright)=lim_{Nto infty}mu([-N,N]cap A)
$$
So, for some large $N'$, $mu([-N',N']cap A)>0$, a contradiction.

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).

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 $f_nto g$ in $L^p,$ then $f=g$ a.e.

Proof: We can find $Asubset mathbb R$ with $m(A^c)=0$ such that $f_nto f$ pointwise on $A.$ Recall that $f_nto g$ in $L^p$ implies $f_{n_k}to g$ pointwise a.e. for some subsequence $f_{n_k}.$ Thus there is a set $Bsubset mathbb R$ with $m(B^c)=0$ such that $f_{n_k}to g$ pointwise on $B.$ On the set $Acap B,$ we have pointwise convergence of $f_{n_k}$ to both $f,g.$ Thus $f=g$ on $Acap B.$ Since $m[(Acap B)^c] = 0,$ we're done.