If a Borel measurable nonnegative $f$ satisfies $\int_{E}fdm<\infty$ for Lebesgue measurable sets $E$ with $m(E)<1$, then $\lim_{n\to \infty}n\cdot m(\{f\ge n\})=0$.
I saw similar questions here, but they assumed that either $\int_{X}f<\infty$ or that $|\int_{E}f|<1$ for $m(E)<1$.
Here, however, the problem being that $\int_{E}fdm$ is finite but can be as big as it can, I am not sure if the same arguments work. Following @Matija's reasoning, I let $g(x)=n\chi_{A}$, where $A={x:f(x)=\infty}$, which means $g_n(x)\le f(x)$. Hence: $\int_{E}n\chi_{A}dm\le \int_{E}{}fdm<\infty$ so $\le n.m(A\cap E)$, but I don't know how to keep from here. This is a bit confusing, and dominated convergence dont seem to apply leaving it in that form. Any suggestions?
Best Answer
First of all observe that $|\{f=\infty\}|=0$. If it has a positive measure, you can find a subset of measure less than $1$ and integrate over it.
We'll now argue by contradiction. Suppose $\limsup n|\{f\ge n\}|=2\delta>0$. Therefore, there exists a sequence $n_k\to \infty$ such that $|\{f\ge n_k\}|\ge \frac{\delta}{n_k}$. After passing to a subsequence we may assume that $|\{n_k\le f\le n_{k+1}\}|\ge \frac{\delta}{2n_k}$ and moreover, we can also assume that $\sum_{k=1}^{\infty} \frac{1}{n_k}<\frac{10}{\delta}$.
For each $k$, choose a subset $B_{n_k}\subseteq \{n_k\le f\le n_{k+1}\}$ such that $\frac{\delta}{20n_k}\leq |B_{n_k}|\le \frac{\delta}{10n_k}$. Define $E:=\cup_{k=1}^{\infty}B_{n_k}$ and note that $|E|<1$. But, we have $$\int_E f\;dm \geq \frac{\delta}{20}\sum_{k=1}^{\infty}\frac{1}{n_k}n_k=\infty.$$ This contradicts the assumption on $f$ and hence we conclude the result!