Weak convergence implies almost everywhere? Suppose that $L^2(R^n;dx) \ni f_n\geq 0$ and $\int f_ng\to 0$ for all continuous function $g$ with compact  support. Then, does $f_n$ convergence to $0$ almost everywhere?
Intuitively, it convergence to $0$ by taking function $g$ s.t., $g=1$ around $x$ for almost every $x$. However I cannot prove.
 A: Another example,  taking place in $L^2[0, 1]$,  which is perhaps easier to understand from a geometric standpoint, is to choose the first 10 functions
in the sequence to be the charachteristic functions of  intervals of length 1/10, sliding from left to right, starting
with  $[0.0,0.1]$  and ending with $[0.9,1.0]$.
Then do the same for the next 100 functions with intervals from
$[0.00,0.01]$  to $[0.99,1.00]$, and so on.
This sequence does not converge pointwise anywhere, but still it converges weakly!
A: As the example by Ruy shows, you may not have pointwise convergence a.e. However, you can get a subsequence which converges a.e.:
Let $B_m$ denote the ball with radius $n$. Due to $f_n \ge 0$, we have
$$\|f_n\|_{L^1(B_m)} = \int_{B_m} f_n \, \mathrm d x = \int_{\mathbb R^d} \chi_{B_m} \, f_n \, \mathrm d x \to 0$$
as $n \to \infty$.
Note that $\chi_{B_m} \in L^2(\mathbb R^d)$ since $B_m$ has finite measure.
By weak convergence, $\|f_n\|_{L^2(B_m)}$ is bounded w.r.t. $n$. Using interpolation,
we get $\|f_n\|_{L^p(B_m)} \to 0$ as $n \to \infty$, where $p \in (1,2)$ is arbitrary (and fixed).
This implies the existence of a subsequence of $f_n$ which converges pointwise a.e. on $B_m$ for any fixed $m$.
A standard diagonal sequence argument yields a subsequence $f_{n_k}$ which converges pointwise a.e. on $B_m$ for all $m$.
From $\mathbb R^d = \bigcup_{m = 1}^\infty B_m$,
$f_{n_k}$ converges pointwise a.e. on $\mathbb R^d$.
