# Sequence of functions divided by constants converge to zero

I am trying to show the following:

If $\{ f_n \}$ is a sequence of a.e. real-valued measurable functions in X, and the measure $\mu(X) < \infty$, there exist positive constants $a_n$ such that $\{ f_n / a_n \}$ converges a.e. to zero.

Here is my attempt: if $f_n$ is a.e. real-valued then the set on which it is not real-valued has measure zero. That set can be covered by some collection of open sets with arbitrarily small positive measure (can it?). Let $E_n$ be such an open covering. Then $f_n$ is everywhere real-valued on $X-E_n$. Furthermore, $X-E_n$ is a closed set, so it contains all its limit points. Therefore $f_n$ assumes a maximum value on $X-E_n$ (does it?), so we can choose a constant $a_n$ such that $|f_n| / a_n$ can be made arbitrarily small. Now if, for example, we let $a_n$ be $n$ times the maximum value of $f_n$ and let $E_n$ have measure $2^{-n}$ this sequence converges a.e. to zero.

I feel like there are a bunch of things wrong with this. I think the gist of the proof is that if the function is a.e. real-valued then we can remove the pieces where it isn't real-valued, and then the function will be bounded everywhere else. That's what I'm trying to show, but I'm not sure how. Any hints?

• In general, $f_n$ does not attain its maximum on $X-E_n$. Simply consider $f_n(x)=x$; then $E_n = \emptyset$ and $f_n$ does not attain its maximum on $\mathbb{R} = \mathbb{R}-E_n$. – saz Oct 13 '14 at 14:19
• @saz Hmm, that's true. Can I somehow use the fact that the measure of $X$ is known to be finite, then? – Andrea Oct 13 '14 at 18:27

Regarding your own thoughts: First of all, if $f_n$ is real-valued, then we can choose a set $F_n$, $\mu(F_n) =0$, such that $\{|f_n|=\infty\} \subseteq F_n$. Since $F:= \bigcup_n F_n$ is still a set of measure zero, we can assume without loss of generality that $|f_n|<\infty$ for all $n \in \mathbb{N}$.
As I already pointed out in my comment, $f_n$ does in general not attain its maximum on $X-E_n$, so your argumentation doesn't work.
1. Show that $$\mu(|f_n| \geq R) \to 0 \qquad \text{as} \, R \to \infty.$$
2. Deduce from the first step that there exists $R_n$ such that $$\mu\left(|f_n| \geq \frac{R_n}{n} \right) \leq \frac{1}{2^n}.$$ Conclude from $$\sum_{n \geq 1} \mu(|f_n| \geq R_n)$$ and $\mu(X)<\infty$ that $$\frac{|f_n(x)|}{R_n} \leq \frac{1}{n}$$ for $\mu$-almost all $x$ and $n=n(x)$ sufficiently large.