Convergence of integrals: if $g_n\ge 0$ and $\int_a^b g_n\to 0$, then $\int_a^b fg_n\to 0$ 
Suppose that $g_n \geqslant  0$ is a sequence of integrable functions which satisfies $\lim\limits_{n \to \infty} \int_a^b g_n(x) \mathrm{d} x = 0$. 
a) Show that if $f$ is an integrable function on $[a,b]$, then $\lim_{n\to \infty} \int_a^b f(x) g_n(x) \mathrm{d}x = 0$.  
b) Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n\to \infty} \int_0^1 x^n f(x) \mathrm{d}x = 0$.

Here is what I have so far;
We are given that $g_n \geqslant 0$, with $\lim_{n\to\infty}\int^b_ag_n(x)=0$. We also have that $f$ is integrable on $[a,b]$. Then, by general form of Mean Value Theorem for Integrals (which I already proved in another problem), we know $\exists c\in[a,b]$ such that $\int^b_af(x)g_n(x)dx=f(c)\int^b_ag_n(x)dx$. Thus we have:
$$\lim_{n\to\infty}\int^b_af(x)g_n(x)dx\Rightarrow
\lim_{n\to\infty}f(c)\int^b_ag_n(x)dx\Rightarrow
f(c)\lim_{n\to\infty}\int^b_ag_n(x)\Rightarrow
f(c)⋅0=0.
$$
(b) We are given that $f$ is integrable on $[0,1]$. Let $g_n(x)=x^n$. Our claim, is that
$$g_n(x)\to g(x)=\left\{\begin{array}{ccc}0&,&x\ne1\\1&,&x=1\end{array}\right.$$
on $[0,1]$.
Then let $x_0\in [0,1)$.  Then $\lim_{n\to\infty}x^n_0=0$.  Thus $g_n(x)\to g(x)$ pointwise on $[0,1]$.  Further  $\lim_{n\to\infty}\int^b_ag_n(x)=\int^b_ag(x)=0$.  Thus, by (a), we have that $\lim_{n\to\infty}\int^b_ax^nf(x)dx=0$.
Is this correct?

The statement of (a) is wrong, you can let 
$$
g_{n}(x)=\begin{cases}
\frac{1}{n\sqrt{x}} & x\in(0,1]\\
0 & x=0
\end{cases}
$$
 and $f(x)=g_1(x)$. Then $\int_0^1g_n(x)dx=\frac{2}{n}\rightarrow 0$ as $n\rightarrow\infty$ and 
$$\int_0^1 f(x)g_n(x)dx=\infty $$
for all $n\in\mathbb{N}$.
