# 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?

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Please use LaTeX. You can use my edits as an example. –  Sasha Aug 6 '12 at 20:44
(b) is pretty much a free consequence of (a) since you can compute $\int_0^1 x^n \, dx$ directly. –  Jason Knapp Aug 6 '12 at 21:00

For part (a), your answer has an error. You can only apply the mean value theorem for integrals if $f$ is continuous. I would have suggested using Cauchy's inequality, but $f$ is not necessarily in $L^2$. Perhaps look at simple functions.
Your answer to (b) is essentially correct, but you should justify moving the limit inside the integral sign. Or just compute the integrals of $g_n$ directly.
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 $n\in\mathbb{N}$.