showing that $\lim_{n\rightarrow \infty} \int_0^1 f_n(x)g(x)~dx =0$ 
Let $\{f_n\}$ be a sequence of functions in $L^2[0,1]$ such that $f_n(x)\rightarrow 0$ almost everywhere on $[0,1]$. If $|f_n(x)|\le 1$ for all $n$ and almost all $x\in [0,1]$.    

I require direction in answering the following questions:  
(a) $$\lim_{n\rightarrow \infty} \int_0^1 f_n(x)g(x)~dx =0$$ for all $g\in L^2[0,1]$.  
(b) Will the conclusion still hold if the condition $|f_n(x)|\le 1$ is replaced by the condition that $||f_n||_{L^2} $ are uniformly bounded?  
Thanks for your responses.
 A: For (a), given any $g\in L^2[0,1]$, since $|f_n(x)|\le 1$ a.e. on $[0,1]$ for all $n$, we have
$$|f_n(x)g(x)|\le |g(x)| \mbox{ a.e. on $[0,1]$ for all }n.$$
Since $g\in L^2[0,1]$, by dominated convergence theorem DCT, we have
$$\lim_{n\rightarrow \infty} \int_0^1 f_n(x)g(x)~dx =\int_0^1 \lim_{n\rightarrow \infty}\big(f_n(x)g(x)\big)~dx=0,$$
where the last equality follows from the assumption that $f_n(x)\rightarrow 0$ a.e. on $[0,1]$ as $n\rightarrow\infty$.
For (b), we can see that the conclusion does not hold if we just assume that $||f_n||_{L^2}$ is uniformly bounded by the following example: take $f_n$ to be 
$$f_n(x)=\left\{
  \begin{array}{ll}
    n, & \hbox{if } x\in[0,\frac{1}{n}]; \\
    0, & \hbox{otherwise.}
  \end{array}
\right.$$
It's clear that $f_n(x)\rightarrow 0$ a.e. on $[0,1]$ as $n\rightarrow\infty$, and $\|f_n\|_{L^2}=1$ for all $n$. Take $g(x)=x^{-\frac{1}{3}}$. Then $g\in L^2[0,1]$ because $\int_0^1g(x)^2dx=\int_0^1x^{-\frac{2}{3}}=3x^{\frac{1}{3}}\Big|^1_0=3$. However, 
$$\int_0^1f_n(x)g(x)dx=n\int_0^{\frac{1}{n}}x^{-\frac{1}{3}}dx=\frac{3}{2}nx^{\frac{2}{3}}\Big|^{\frac{1}{n}}_0=\frac{3}{2}n^{\frac{1}{3}}\rightarrow\infty$$
as $n\rightarrow\infty$.
