# For $\mu(X)<\infty$ and $f_n<\infty$ a.e. s.t. $\forall M$ $\exists n$ s.t. $\mu(\{f_n>M\})\geq\delta$, then $f_n$ diverges on set of positive measure

Let $f_n$ be a sequence of measurable finite a.e. functions on a space $X$ of finite measure. Given $\delta >0$, suppose that for all $M>0$ there exists an $n$ such that $\mu(\{f_n>M\})\geq\delta$. Prove that $f_n$ is unbounded on a set of positive measure.

I'm looking for a clean way to do this that makes it appear as obviously true as possible. The best I could come up with was to set $M=\left \lceil \frac{\mu(X)}{\delta} \right \rceil$, to figure out how many disjoint sets of measure $\delta$ it would take to sum up to the measure of $X$. And then look at the collection

$$\{f_{n_1}>1\},\{f_{n_2}>2\},...,\{f_{n_M}>M\},\{f_{n_{M+1}}>M+1\}$$

And conclude that these sets cannot be pairwise disjoint modulo a set of measure zero. And then keep repeating the process again and again to find a subsequence $f_{n_k}$ and a descending sequence $E_k$ of sets of positive measure on which the subsequence $f_{n_k}$ increases without bound.

Assuming this is even correct, is there a simpler way to solve this?

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For an integer $M$, let $A_M = \bigcup_{n=1}^\infty \{f_n > M\}$. Since $\mu(\{f_n > M\}) \ge \delta$ for at least one $n$, we have $\mu(A_M) \ge \delta$. Moreover, it should be clear that $A_1 \supset A_2 \supset \dots$. Set $A = \bigcap_M A_M$; by continuity from above (which holds because our measure is finite), we have $\mu(A) = \lim_{M \to \infty} \mu(A_M) \ge \delta$. But for $x \in A$, for every $M$ there is an $n$ such that $f_n(x) > M$, which is to say that $\{f_n(x)\}$ is unbounded.
As an alternative formulation: suppose $f$ is a measurable function with $f < \infty$ almost everywhere. Then, again using continuity from above, $0 = \mu(f = \infty) = \lim_{M \to \infty} \mu(\{f > M\})$. So for any $\delta > 0$, there exists $M$ such that $\mu(\{f > M\}) < \delta$. Take $f = \sup f_n$ and you will see this is precisely the contrapositive of the desired statement.
Brilliant, exactly what I was looking for, I knew there had to be some theorem which used the fact that $X$ was finite, and there it was, descending continuity of measure, hiding right under my nose the whole time. Thanks. – esproff May 29 '13 at 4:27