$u_n$ weakly converges to $u$ in $H_0^1(I)$, $u_n'$ converges pointwisely to $u'$ I is a bounded interval, like (0,1). 
$u_n$ is a sequence weakly converges to $u$ in $H_0^1(I)$, do we have $u_n'$ converges pointwisely to $u'$? If it is not true, can we pass it to a subsequence and then have the pointwise convergence?
Moreover, if we have the pointwise convergence of $u_n'$, do we have even better convergence, like $L^2(I)$ convergence?
 A: No.
A complex example is a little easier to give, so let's take $I = (0, 2\pi)$ and $f_n(x) = \frac{1}{n} (e^{i n x}-1)$, so that $f_n'(x) = ie^{i n x}$. Then $f_n$ converges weakly to zero.   This follows from the fact that $f_n \to 0$ strongly in $L^2$, and $f_n' \to 0$ weakly in $L^2$ (the $f_n'$ are, after rescaling by $2\pi$, an orthonormal sequence, so Bessel's inequality implies that they converge weakly to zero.)  So if you write down the inner product of $f_n$ with any $g \in H^1_0(I)$, you can easily show it converges to zero.
Now you can see directly that $f_n'(x)$ diverges for every $x \in (0,2\pi)$.  If you pass to a subsequence, you can get it to converge at a few points, but not a.e.  For suppose there were some subsequence $f_{n_k}'(x) = ie^{i n_k x}$ converging almost everywhere to some function $h$.  Then by dominated convergence, it would also converge to $h$ strongly in $L^2$.  Now since $f_{n_k}' \to 0$ weakly in $L^2$ as we saw above, this would imply $f_{n_k}' \to 0$ strongly in $L^2$.  But this is absurd because $\|f_{n_k}'\|_{L_2} = 2\pi$ for every $k$.
If you want an example over $\mathbb{R}$, take imaginary parts, so $f_n(x) = \frac{1}{n} \sin (nx)$ and $f_n'(x) = \cos(nx)$.  (Real parts would also work.)  
I think it's true that any subsequence $f_{n_k}'$ actually diverges almost everywhere, but I don't have a proof off the top of my head.  
