Let's build a "fat Cantor set".  Start with $A_0 = [0,1]$ with measure $\alpha_0=1$.
Then remove a short open interval centered at $1/2$, leaving a set $A_1 \subset A_0$ of measure $\alpha_1 < \alpha_0$.  So $A_1$ is made up of $2$ closed intervals of length $\alpha_1/2$.   Let $B_1$ be the removed interval with length $1-\alpha_1 = \alpha_0-\alpha_1$.  

Next remove a short open interval from the center of each of the intervals of $A_1$, to leave $A_2 \subset A_1$ of measure $\alpha_2<\alpha_1$.  And $A_2$ is made up of $4$ closed intervals of length $\alpha_2/4$.  Let $B_2$ be made up of the $2$ removed intervals, each of length $(\alpha_1-\alpha_2)/2$.

Continue in this way.  $A_n \subset A_{n-1}$ has measure $\alpha_n < \alpha_{n-1}$, and $A_n$ is made up of $2^n$ closed intervals each of length $\alpha_n/2^n$.  $B_n$ consists of the $2^{n-1}$ newly removed open intervals, each of length $(\alpha_n-\alpha_{n-1})/2^{n-1}$

Let $A = \bigcap_{n=1}^\infty A_n$.  Choose the lengths of the intervals removed so that $\alpha>0$, where $\alpha = \lim_{n \to \infty} \alpha_n$.  (This is what makes it a "fat" Cantor set.)  Of course $m(A) = \lim_{n \to \infty} m(A_n) = \alpha > 0$, where $m$ is Lebesgue measure.  

Our limit function is
$$
f = \frac{1}{\alpha} \mathbf1_A
$$
where $\mathbf1_A$ denotes the indicator function of set $A$.  For $n\ge 1$ define
$$
f_n = \frac{1}{(\alpha_n-\alpha_{n-1})}\mathbf1_{B_n}
$$
(I used Bill Johnson's idea of making an $l^1$ basis.  But now these really are disjoint, so you don't have to do estimates to show they are "close enough" to being disjoint.)
Now we claim: 
>(1) $\int f_n g$ converges to $\int f g$ for all continuous $g$; 

>(2) there is $h \in L^\infty[0,1]$ such that $\int f_n h$ does not converge to $\int fh$.  



**(1)**  

Let $g$ be a continuous function.
Let $\pi_n$ be the partition of $[0,1]$ made up of the $2^{n+1}$ endpoints of the set $A_n$.  Note that for each interval $I$ of partition $\pi_n$ we have
$$
\int_I f_{n+1} = \int_I f
$$
and more generally
$$
\int_I f_k = \int_I f
$$
for all $k > n$.  As $n \to \infty$, the lengths of these intervals goes to $0$.  And $g$ is uniformly continuous.  So we will conclude that $\int f_n g \to \int f g$.

**(2)**  

Let $h = \mathbb1_A$.  Then $f_nh=0$ so $\int f_nh = 0$ for all $n$.  But $fh=f$ a.e. and $\int fh = \int f = 1$.