Consider the measure space $(\mathbb{Z},\mathcal{P}(\mathbb{Z}),\#)$, where $\#$ is the counting measure on $\mathbb{Z}$ and $\mathcal{P}(\mathbb{Z})$ is its power set.
I would like to show that for any measurable function we have $\int f(n)d\#(n)=\sum_{n}f(n)$.
This is what I have done: Let $x\in\mathbb{Z}$ and consider the indicator function $1_{\{x\}}$. Then $$\int_\mathbb{Z} fd\#=\int_\mathbb{Z} 1_{\{x\}}d\#=\#\{x\}=1,$$ for $f=1_{\{x\}}$. Next, for a step function $f=\sum_{k=-n}^na_k1_{\{x_k\}}$ (where $x_k\in\mathbb{Z}$ and $a_k$ are real rumbers for all $k$) we have $$\int_\mathbb{Z} fd\#=\sum_{k=-n}^na_k\int_\mathbb{Z}1_{\{x\}}d\#=\sum_{k=-n}^na_k.$$
How do I finish this proof? I still need to prove the statement for an arbitrarily measurable function.
Best Answer
I think the statement holds only for positive measurable functions. If you take $ f(n) = \frac{(-1)^n}{n} $ then $ \sum_n f(n) < \infty $ but $$ \int_\mathbb{Z} f^+ d\# = \int_\mathbb{Z} f^- d\# = \infty $$ So you end up with $\infty-\infty $