Where $f:[0,1]\to\mathbb R$ is Lebesgue integrable, show $\lim_{n\to\infty} n\lambda(\{x:|f(x)|\geq n\})=0$.

Question

Title says it all. It's clear why $\lim_{n\to\infty}\lambda(\{x:|f(x)|\geq n\})=0$ -- since otherwise for arbitrarily high $n$ there'd be a subset of $[0,1]$ with nonzero measure where $|f|\geq n$ and thus the integral would not be finite. But when we include the multiplication by $n$, I'm at a loss.

Proof by reductio seems a poor strategy here, but I don't know any theorems that would seem to apply to this situation. Any pointers would be appreciated.

Answer 1

Use dominated convergence on the functions $n 1_{\{\vert f \vert\geq n\}}$. The pointwise limit is zero and they are pointwise bounded by $f$.

Answer 2

Try showing that $\lim_{n\to\infty} \int_{\{x:\vert f(x)\vert \geq n\}} \vert f\vert=0$ using the Dominated Convergence Theorem. In particular, think about what the sequence of functions is here, what it is dominated by, what this sequence converges to pointwise almost everywhere etc. then relate your value to that integral.