Problem:
Show that for any $g\in L^1([0,\infty))$ we have that $\lim_{n\to\infty} \frac{1}{n}\int_0^n xg(x) dx = 0$.
Attempt:
I haven't been able to do much, but I know that we can say that: $\frac{1}{n}\int_0^n xg(x) dx =\frac{1}{n}\int_0^\infty xg(x)\chi_{[0,n]} dx$, where $\chi_{[0,n]}$ is the indicator function for the interval $[0,n]$. I believe it's better to write the integral like this because then the bounds are not "moving", which usually helps in my experience. I think that considering the functions $h_n(x)=\frac{1}{n}xg(x)\chi_{[0,n]}$ might be useful. We also have that $\int_0^\infty |g|<\infty$. I believe that we ought to find a function that dominates the $h_n$'s that has the form of $\frac{1}{n}\triangle$ where $\triangle$ is finite as then the desired limit would follow. If instead of $\frac{1}{n}$ in the limit we had $\frac{1}{n^2}$ it would be easy...
$|h_n(x)| \leq |g(x)|$ for all $X$ and for all $n$. Since $h_n(x) \to 0$ for each $x$ the integral tends to $0$ by Dominated Convergence Theorem.