I'm trying solve this but I'm not sure that is correct.
The exercise is:
If $f:[0,1]\rightarrow\overline{\mathbb{R}}$ is lebesgue integrable, show that:
$$ \lim_{n\rightarrow+\infty}\int_0^1x^nf(x)=0 $$
My attempt:
Consider the sequence $\{|x^nf(x)|\}$ that are a sequence uniformly limited by $|f(x)|$ (this is it? I can affirm this?). Note that: $|x^nf(x)|\leq|f(x)|$. As $f(x)$ is lebesgue integrable so $|f(x)|$ is too. Now I apply the Dominated convergence theorem but I do not know if that's … I can say that this sequence converges punctually?
Best Answer
You've got pretty much all the pieces already:
Since $f$ is integrable, it's finite almost everywhere. Hence if $0 \le x < 1$, and $f(x)$ is finite, $$\lim_{n \to \infty} x^n f(x) = f(x) \lim_{n \to \infty} x^n = 0$$
Furthermore, $x^n \le 1$ for all $x \in [0, 1]$, so that $|x^n f(x)| \le |f(x)|$ for a.e. $x$. Thus the dominated convergence theorem applies and
$$\lim_{n \to \infty} \int_0^1 x^n f(x) dx = \int_0^1 \lim_{n \to \infty} x^n f(x) dx = \int_0^1 0 dx = 0$$