Let $f:\Bbb R\to \Bbb R$ be defined by
$f(x)$=\begin{cases} \frac{1}{\sqrt x};\text{$0
Define $g:\Bbb R\to \Bbb R$ by $g(x)=\sum_{n=1}^\infty \dfrac{1}{2^n}f(x-r_n)$ where $\{r_n\}_n$ is an enumeration of rationals in $\Bbb R$.
Show that $f,g$ is integrable.
Since $f$ is unbounded we need to talk about Lebesgue-Integrability here.
Now $\int f=2\sqrt x|_0^1=2<\infty \implies f$ is Lebesgue Integrable.
I am unable to show this for $g$. Any help.