Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
How did you define the Lebesgue integral? –  Michael Greinecker Dec 5 '12 at 9:40
2  
And how did you define infinite sums over $\mathbb{Z}$? –  Michael Greinecker Dec 5 '12 at 9:45

2 Answers 2

Hint: Let $f$ be a measurable function, $f \geq 0$. Then there exists a sequence $(f_n)_n$ of step functions such that $f = \sup_n f_n$. Now apply monotone convergence and use the formula for the step functions (which you already proved).

If $f$ is an arbritary measurable function you can write $f$ as $f=f^+-f^-$ where $f^+$, $f^- \geq 0$ are measurable functions.

share|improve this 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 $

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.