If $f\in L^1([0,\infty ))$ then $\lim_{n\to \infty}\frac{1}{n}\int_{[0,n]}xf(x)dx=0$

Question

Let $f$ be a Lebesgue measurable function such that $\int_{[0,\infty]}\left |f(x)\right |dx<+\infty$ (Lebesgue's integral). Prove that $\lim_{n\to \infty}\frac{1}{n}\int_{[0,n]}xf(x)dx=0$

Answer 1

Note that for every $x \in [0,\infty)$, we have $$ \underbrace{\left|\frac{xf(x)\mathcal{X}_{[0,n]}}{n}\right|}_{\triangleq f_n(x)}\leq |f(x)| $$

and $f_n(x) \to 0$ as $n\to\infty$ (as for every given $x$, $\frac{xf(x)}{n}\to 0$ as $n\to\infty$). Finally, since $f \in L^1$, we have by dominated convergence theorem that $$ 0 = \lim_{n\to\infty}\int_{[0,\infty)}\frac{1}{n}xf(x)\mathcal{X}_{[0,n]}dx = \lim_{n\to\infty}\frac{1}{n}\int_{[0,n]}xf(x)dx. $$