Lebesgue integral convergence

Question

Suppose that $f : [0, 1] \to \mathbb{R}$ is Lebesgue integrable. Show that $x^n f(x)$ is also Lebesgue integrable on $[0, 1]$ for every $n \geq 1$ and compute $\lim_{n \to \infty} \int_{[0, 1]} x^n f(x)$.

I was wondering if I could get a hint.

Answer 1

To show the first part, finding an integrable bound for a $\left| x^nf(x)\right|$ will work just fine, as $x^nf(x)$ is a product of Lebesgue measurable functions and if $g\colon [0,1]\to \mathbb{R}$ is an integrable function s.t. $\left| x^nf(x)\right| \le g(x)$ a.e. then $$\int_{[0,1]}\left| x^nf(x)\right|\le\int_{[0,1]}g(x)\lt\infty.$$ Once you find an integrable bound (which is really straightforward here), you can apply Lebesgue's dominated convergence theorem, so $$\lim\limits_{n\to\infty}\int_{[0,1]}x^nf(x)\,\text{d}x=\int_{[0,1]}\lim\limits_{n\to\infty}(x^nf(x))\,\text{d}x$$ will hold. Obviously, on the right hand side there is a.e. pointwise limit.