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

Question

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?

Answer 1

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. $x\in (0,1)$.

Answer 2

So $a_{n}\leq 1$ for all $n$. Consider $\displaystyle\sum_{n}\int_{0}^{1}|f(a_{n}x)|\dfrac{dx}{x^{2}}$, a change of variable $y=a_{n}x$ leads to $\displaystyle\sum_{n}a_{n}\int_{0}^{a_{n}}|f(y)|\dfrac{dy}{y^{2}}\leq\displaystyle\sum_{n}a_{n}\int_{0}^{1}|f(y)|\dfrac{dy}{y^{2}}<\infty$, but $\displaystyle\sum_{n}\int_{0}^{1}|f(a_{n}x)|\dfrac{dx}{x^{2}}=\displaystyle\int_{0}^{1}\sum_{n}|f(a_{n}x)|\dfrac{dx}{x^{2}}$, so $\displaystyle\sum_{n}|f(a_{n}x)|<\infty$ a.e. $\dfrac{dx}{x^{2}}$ on $(0,1)$, it also implies a.e. on $(0,1)$ since for measurable $A\subseteq (0,1)$, we have for Lebesgue measure $\mu$, $\mu(A)=\displaystyle\int_{0}^{1}\chi_{A}dx=\int_{0}^{1}x^{2}\chi_{A}\dfrac{dx}{x^{2}}\leq\int_{0}^{1}\chi_{A}\dfrac{dx}{x^{2}}$.