Does almost everywhere convergence imply convergence in $L^p$? Is it true that if $\Omega$ is an open, bounded subset of $R^{N}$, $u_{n} \to u$ almost everywhere, $1<p<\infty$, then $\|u_{n} - u\|_{L^{p}} \to 0$? Here sequence and the function are in $L^{p}$
My proof uses dominated convergence theorem to $v_{n}(x) = |u_{n}(x) - u(x)|^{p}.$
and the fact that $\Omega$ is bounded.   
thanks
 A: For $\Omega=(0,2)$ and $N=1$,
take $u_n=n\chi_{(0,1/n]}$. Then $u_n$ converges to $0$ almost everywhere, but
$  ||u_n-0||_p=n \cdot (1/n)^{1/p}=n^{1-{1\over p}}$ for all $n$.
Similar examples can be constructed for other $\Omega$.
A: I will give a counter-example for an arbitrary open $\Omega$. Take $x_0\in\Omega$; for $n$ large enough $B(x_0,n^{-1})\subset \Omega$. Put $f_n(x):=n^{\frac Np+1}\mathbf 1_{B(x_0,n^{-1})}(x)$. $f_n(x)$ converges to $0$ except at $x_0$ which has $0$ measure, but
$$\lVert f_n\rVert_p=n^{\frac Np+1}m(B(x_0,n^{-1}))^{\frac 1p}=n^{\frac Np+1}\left(\frac 1{n^N}\right)^{\frac 1p}v_N=nv_N,$$
where $v_n$ is the measure of the unit ball of $\mathbb R^N$,
so we can 't have the $L^p$ convergence. 
But there is a link between $L^p$ and almost everywhere convergences: if $\{f_n\}\subset L^p$ is a sequence which converges in $L^p$, we can extract an almost everywhere converging subsequence.
Your proof fails because in general we can't find an integrable function $g$ such that $|f_n|\leq g$ for all $n$. In my example, the we can try $g(x)=\sum_{n\geq 1}n^{\frac Np+1}\mathbf 1_{B(x_0,n^{-1})\setminus B(x_0,(n+1)^{-1})}$, but 
$$\int_{\Omega}|g¬^pd\lambda =\sum_{n\geq 1}n^{N+p}v_N\left(\frac 1{n^N}-\frac 1{(n+1)^N}\right)=v_N\sum_{n\geq 1}n^p\left(1-\frac 1n\right),$$
which is not convergent. 
When you try to bound $|u_n-u|$, there is a problem, since the "$n$ large enough" depends on the point $x$. It's possible we can't take the same for almost every $x$.
