Here's another question that I'm stuck on from my studies for an upcoming exam. This one comes from another practice preliminary exam.
Problem
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a Lebesgue integrable function. Prove that for almost every $x \in \mathbb{R}$ that $$ \lim_{n \rightarrow \infty} f(x) f(2^2x) f(3^2x) \cdots f(n^2x) = 0$$
I.e. given that $\int_\mathbb{R} |f(x)| dx < \infty$, prove the above limit.
Please, correct me if I'm wrong, confirm my thoughts, or provide hints towards a solution. Thanks in advance!
My Partial Attempt
Break the support into three cases. Either $S := \{ x \in \mathbb{R} | f(x) \neq 0\}$ (1) has $\mu(S) = 0$, (2) finite $\mu(S) > 0$, or (3) $\mu(S) = \infty$,.
(1) We have immediately that the integral $\int_\mathbb{R} |f(x)| dx = 0$ because the function is supported by a set of measure zero. Hence, the function is zero almost everywhere and we have our result.
(2) This is the tricky case for me. This is where I'm looking for guidance if my case-by-case method is in fact a method for solution. Otherwise, provide a hint or an alternative solution method.
(3) Help here too.
For $j=1,2,\dots$ and $\epsilon>0$, define
$$A_j^\epsilon = \{ x: |f(j^2 x)|>\epsilon\}.$$
This is a measurable set. We know
$$ \epsilon m (A_j ^\epsilon) \le \int |f |(j^2 x ) dx = \frac{\int |f| (y) dy}{j^2}=\frac{c}{j^2}.$$
Therefore $m (A_j ^\epsilon) \le \frac{c}{j^2\epsilon}$.
Write $A_j = A_j^{1/\sqrt{j}}$. Then $m (A_j ) \le \frac{c}{j^{3/2}}.$
Now let $A=\limsup_{j} A_j = \{x: |f(j^2 x) |\ge \frac{1}{\sqrt{j}}\mbox{ for infinitely many }j\mbox{'s}\}$
It follows from the Borel-Cantelli lemma for general measure spaces that
$$ m (A)=0.$$
On the complement of $A$
$$|f(j^2 x)| \le \frac{1}{\sqrt{j}},$$
for all $j$ large enough (depending on $x$), and the result follows.