$\lim_{n\rightarrow\infty}\int_0^1 f(x)g_n(x)dx=f(0)$ Let $f: [0,1]\rightarrow\mathbb{R}$ be continuous. Define $g_n$ as $g_n=n\lambda_{[0,1/n]}$ with $n\geq 1$ ($n$ times the indicator function of $[0,1/n]$. Show that $\lim_{n\rightarrow\infty}\int_0^1 f(x)g_n(x)dx=f(0)$. 
I tried to use convergence theorem, but it certainly is not monotone converging, and I can't find a way to use Lebesgue Convergence theorem... I'm not quite sure what $fg_n$ converges to...
 A: Since $f$ is continuous, given $\epsilon>0$, we have $|f(x)-f(0)|<\epsilon$ for $x\in[0,1/n]$ and all sufficiently large $n$. Hence, since $\int_0^1g_n=1$ we have
$$
\left|\int_0^1fg_n-f(0)\right|=\left|\int_0^1(f-f(0))g_n\right|\le\epsilon\int_0^1g_n=\epsilon.
$$
PS: Note that
$$
\int_0^1g_n=\int_0^1 n\chi_{[0,1/N]}=n\int_0^{1/n}1=1
$$
and so
$$
f(0)=f(0)\int_0^1g_n=\int_0^1 f(0)g_n.
$$
A: Denote $f(\alpha_{n})=\min_{x\in[0,1/n]}f(x)$ and $f(\beta_{n})=\max_{x\in[0,1/n]}f(x)$ then taking integral to each side of $f(\alpha_{n})g_{n}(x)\leq f(x)g_{n}(x)\leq f(\beta_{n})g_{n}(x)$ we get 
\begin{align*}
f(\alpha_{n})=n\int_{0}^{1/n}f(\alpha_{n})dx\leq\int_{0}^{1}f(x)g_{n}(x)dx\leq n\int_{0}^{1/n}f(\beta_{n})dx=f(\beta_{n}),
\end{align*}
now $f(\alpha_{n}),f(\beta_{n})\rightarrow f(0)$.
A: By mean value theorem for definite integrals we have to 
 if $f : [a, b] \to \mathbb{R}$ is continuous and $g$ is an integrable function that does not change sign on $[a, b]$, then there exists $c$ in $(a, b)$ such that
$$
\int_a^b f(x) g(x) \, dx = f(c) \int_a^b  g(x) \, dx.
$$
Note that
$$
\int_0^1 f(x) g_n(x) \, dx = \int_0^{1/n} f(x) g_n(x) \, dx
$$
In this question, for all $n$ we have an $c_n\in (0,1/n)$ such that
$$
\lim_{n\to\infty}\int_0^{1/n} f(x) g_n(x) \, dx
 = 
\lim_{n\to\infty} f(c_n) \int_0^{1/n}  g_n(x) \, dx 
=
\lim_{n\to\infty}f(c_n)\cdot n \cdot \int_0^{1/n} \, dx 
=
\lim_{n\to\infty}f(c_n)
=f(0) 
$$
