Let $f:\mathbb R\to [0,\infty]$ be a Lebesgue measurable function which is also Lebesgue integrable. Define, for each positive integer $n$, $S_n=\{x\in \mathbb R:\ f(x)>n\}$. Let $\varepsilon>0$.

I am pondering upon the following:

Does there exist a large enough $N$ such that whenever $n\geq N$, we have $$ \int_{S_n}f\ d\mu < \varepsilon $$

Here's what I am thinking:

Wrtie $X=\mathbb R$. Let $\int_X f\ d\mu =I<\infty$.

Thus we must that $\mu(S_k)<\infty$ for all $k$.

Define $S=\bigcap_{k\in \mathbb N} S_k$.

Since $\mu(S_1)$ is finite, we can write $\lim_{k\to \infty}\mu(S_k)=\mu(S)=\ell$ (say).

Now $\int_Sf\ d\mu\geq n\mu(S)=n\ell$ for all $n$. Thus, if $\ell\neq 0$, then we would have $\int_S f\ d\mu = \infty$, contradicting the integrability of $f$.

So we must have $\ell=0$.

I don't see where to go from here.

  • $\begingroup$ Now estimate $\int_{S_n}f \ge n\mu(S_n)$. $\endgroup$ – GEdgar Mar 10 '15 at 18:43
  • $\begingroup$ @GEdgar Hello Edgar. Can you please elaborate on this. I am a little lost. $\endgroup$ – caffeinemachine Mar 10 '15 at 18:56
  • $\begingroup$ You can also show that $\chi_{S_n} (x) \to 0$ for almost every $x$ (namely, for all $x$ with $f(x) < \infty$) and then use the dominated convergence theorem. $\endgroup$ – PhoemueX Mar 10 '15 at 20:21

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