Proving that given a measure, the function $f \in L^{1}$

Question

I got this problem I cant figure out by the definitions Im given :/. Any hint will be really appreciated!

Let $\Omega=\{1,2,...\}$, $F=2^{\Omega}$ and $\mu$ defined over $(\Omega, F)$ by $\mu(k)=p(1-p)^{1-k}$ for $k\geq 1$.

Let $f:\Omega\rightarrow \mathbb{R}$ defined by $f(k)=k$. Prove that $f\in L^{1}(\Omega,F,\mu)$ and calculate $\int_{\Omega}fd\mu$.

Once $f \in L^{1}$ I think that I can calculate it, but I think I do not have the tools to prove it!

Thanks so much for your advices!

Answer 1

I think you got a typo! cause if $\mu(k)=p(1-p)^{1-k}$ f is not in $L^{1}$.

First, lets notice that $\mu(\Omega)=\sum_{k\in \mathbb{N}} \mu(k)=p\sum_{k\in\mathbb{N}}(1-p)^{k-1}=\frac{p}{1-p}\sum_{k\in\mathbb{N}}(1-p)^{k}=\frac{p}{1-p} \frac{1-p}{1-(1-p)}=1$ for $p\in(0,1)$, $0<(1-p)<1$.

Then, as pointed in the comments, $\int_{\Omega}fd\mu=\int_{\bigcup{\{k\}}}fd\mu=\sum_{k\in\mathbb{N}}\int_{\{k\}}kd\mu=1\mu(1)+2\mu(2)+...=\sum_{k\in\mathbb{N}}k\mu(k)$. This is because each $\int_{A_{k}} cd\mu = c\mu(A_{k})$, $c\in \mathbb{R}$. Now, $\sum_{k\in\mathbb{N}}k\mu(k)=\sum_{k\in\mathbb{N}}kp(1-p)^{k-1}=p\sum_{k\in\mathbb{N}}k(1-p)^{k-1}=p(\sum_{k\in\mathbb{N}}(1-p)^{k})'=p\frac{1}{(1-p)^2}$.

Any corrections are welcome cause im pretty new at this measure theory stuff