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I have learnt the following theorem: Let $(X,A,\mu)$ be a measure space and $f:X \rightarrow \overline{\mathbb{R}}$ a measurable function. Then $f$ is summable if and only if $|f|$ is summable. (Note: by summable I mean that the integral exists and is finite).

I want to apply this theorem to the space $(\mathbb{N}_{>0},2^\mathbb{N}, \#)$ where $\#$ is the counting measure, considering the function $f:\mathbb{N}_{>0}\rightarrow \overline{\mathbb{R}}$ is defined by $f(n)=(-1)^n \frac{1}{n}$. Then I get that $f$ is measurable and the integral equals to the series, but one $\int |f| d\# = \sum_{n=1}^\infty \frac{1}{n} = \infty $ whereas $\int f d\# = \sum_{n=1}^\infty (-1)^n\frac{1}{n} <\infty $. Where am I wrong? Thank you for your help.

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You are mixing up different notions. Convergence of a series $\sum a_n$ does not imply that the function $f(n)=a_n$ is integrable w.r.t. the counting measure. In your example $f$ is not integrable.

Also note that Riemann integrability of $|f|$ does not imply Riemann integrability of $f$ in contrast with measure theoretic notion of integrability.

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