# Show that $f,g$ is integrable.

Let $f:\Bbb R\to \Bbb R$ be defined by

$f(x)$=\begin{cases} \frac{1}{\sqrt x};\text{$0<x<1$}\\ 0;\text{else} \end{cases}

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.

• Why do we need to talk about Lebesgue integrability? $f$ is Riemann Integrable imo. As you did, you took the limit for $r \to 0$ of the Riemann Integrals in $(r,1)$. – Maffred Jan 27 '17 at 3:40
• But $f$ is not bounded ? As $x\to 0\implies f(x)\to \infty$ – Learnmore Jan 27 '17 at 3:48
• I don't think this is a problem becuase you integrate in $[R,1]$ and then take the limit. – Maffred Jan 27 '17 at 4:25
• "Lebesgue's integrability has nothing to do with the Lebesgue integral" cit wikipedia, my bad! :) You can call it this way if you please. – Maffred Jan 27 '17 at 4:29

Use Tonelli: For non negative measurable $g_n$, we have $\int \sum_n g_n = \sum_n \int g_n$, so if we let $g_n(x) = {1 \over 2^n} f(x-r_n)$, we see that $\int g = \int \sum_n g_n = \sum_n \int g_n = \sum_n 2 {1 \over 2^n} < \infty$.