Q:

Explain a step in the proof that if $f(x)/x^2 \in L^1(0,1)$ and $\sum\limits_na_n=1$ then $\sum\limits_nf(a_nx)$ converges almost everywhere

My question comes from, but is not, the solution to this problem: 

Let $\frac{f(x)}{x^2} \in L^1(0,1)$ and $a_1,a_2,...$ be positive
  reals such that $\sum_{n=1}^{\infty} a_n = 1$. Show that
  $$\sum_{n=1}^{\infty} f(a_nx)$$ converges a.e on $(0,1)$.

My professor began by considering the problem under the weighted lebesgue measure 
as 
$$\int_0^1\left|\sum_{n=1}^{\infty} f(a_nx)\right|\frac{dx}{x^2}$$
and showed that this was finite, and then stated that...

...since the result holds for the weighted measure, it must also hold for the standard Lebesgue measure.

I am not really sure how to verify this statement, since I have never seen any result on weighted measures or their relation to the original measure. 

Why exactly is the last highlighted statement above, true? 

A:

Let $\mu$ be the measure defined on Lebesgue-measurable subsets of $(0,1)$ by
$$
\mu(A) := \int_A \frac{1}{x^2}\, dx,
\qquad A\subset (0,1)\ \text{measurable}.
$$ 
(Observe that you may have $\mu(A) = +\infty$.)
Since $m(A) \leq \mu(A)$ (being $m$ the Lebesgue measure), then $\mu(A) = 0$ implies $m(A)=0$.
Hence, if the series converges for $\mu$-a.e. $x\in (0,1)$, then it converges also for $m$-a.e. $x\in (0,1)$. 

