I try to solve this following problem but I don't get any progress in showing $\ g$ to be integrable:
Define $f:\mathbb{R} \to \mathbb{R}$ by $$f(x)=\begin{cases}\frac{1}{\sqrt x}& \text{if}\ 0<x<1 \\0&\text{otherwise} \end{cases}$$ Let $\{r_n\}_{n=1}^{\infty}$ be the enumeration of all the rational numbers. Define $g:\mathbb{R} \to \mathbb{R}$ by $$g(x)=\sum_{n=1}^{\infty}\frac{1}{2^n} f(x-r_n).$$ Prove that $\ f$ and $\ g$ are integrable on $\mathbb{R}$.
The function $\ f$ is continuous except $x=0$ so this is integrable. But what about $\ g$? The functions $f_n(x) :=\frac{1}{2^n}f(x-r_n)$ is basically $0$ outside $(r_n, r_n+1)$.
If I can show that $\ g$ is continuous almost everywhere on $\mathbb{R}$ then done....I have the following question on mind:
Does the sequence of partial sum of the series $\sum f_n$ is uniformly convergent almost everywhere to $\ g$?
If So then I think $\ g$ becomes continuous $a.e$ on $\mathbb{R}$ and hence continuous.
But I can't prove this assertion....If I made any mistake let me know. Thankyou.
Please help.
$\int g(x)dx =\int \sum_{i=1}^{\infty} \frac 1 {2^{n}} f(x-rn)dx=\sum_{i=1}^{\infty}\int \frac 1 {2^{n}} f(x-r_n)dx =\sum_{i=1}^{\infty} \frac 1 {2^{n}}\int f(x)dx=\int f(x)dx=2$. Interchange of sum and integral is justified by Tonelli's Theorem and the third equality follows from the fact $\int_{-\infty}^{\infty} f(x-a)dx =\int_{-\infty}^{\infty} f(x)dx$ for any $a$.