Let $f \in L^1(0,\infty)$ be non-negative. Calculate $\lim_{n \to \infty} \frac{1}{n} \int_{0}^n xf(x)dx$.
$\lim_{n \to \infty} \frac{1}{n} \int_{0}^n xf(x)dx$ = $\lim_{n \to \infty} \int_{0}^\infty \frac{1}{n}xf(x)\chi_{[0,n]} dx$. If we define $g_{n}(x)$:= $\frac{1}{n}xf(x)\chi_{[0,n]}$, then obviously $g_{n}$ $\to$ $0$. How can we find a dominant function for $\frac{1}{n}xf(x)\chi_{[0,n]}$? I though about change of variable by defining $y:=\frac{x}{n}$, but still I couldn't find a dominant function.
Observe that, since $$ 0\leq \frac{x}{n}\chi_{[0,n]}(x) \leq 1, \qquad \forall x\geq 0, $$ one has, for every $n\geq 1$, $$ |g_n(x)| \leq |f(x)|, \qquad \forall x\geq 0. $$