$\lim_{n\rightarrow\infty}\int_0^1 f(x)g_n(x)dx=f(0)$ 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...

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
$$
f(0)=f(0)\int_0^1g_n=\int_0^1 f(0)g_n.
$$
