I am trying to prove the following:
Show that for each $g \in L^1([0,\infty))$,$$\lim_{n \to \infty}\dfrac{1}{n} \int_0^n xg(x)dx=0$$
What I did up to now is:
Let $h(x)=xg(x)$, then $|h| \leq n|g|$ in $[0,\infty)$, it follows $h \in L^1([0,\infty))$, then $$\dfrac{1}{n} \int_0^n xg(x)dx= \int_0^n \dfrac{1}{n}xg(x)dx$$
If I call $g_n(x)=\dfrac{1}{n}xg(x)$, we have $g_n \to 0$
I don't know what to do from here, any suggestions would be appreciated.
$\displaystyle \int_0^n \frac{1}{n}xg(x) dx= \int_0^{\infty} \frac{x}{n} \chi_{[0,n]}g(x)$
But $\displaystyle |\frac{x}{n} \chi_{[0,n]}g(x)| \leq |g(x)|$ for all $n$. The result follows from the dominated convergence theorem.