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If you consider the measure space $(\mathbb{N},P(\mathbb{N}),\mu)$, where $\mu$ is the counting measure, and if $f:\mathbb{N}\to [0,\infty]$ is measurable, then lebesgue integration with respect to $\mu$ corresponds to a series.

What if instead $f:\mathbb{N}\to\mathbb{C}$ and $f$ is integrable. Does it follow that $\int_{\mathbb{N}}fd\mu=\sum_nf(n)$, and if so why?

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    Hint: dominated convergence. – 2017-01-24
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    How do you define the integral of a complex-valued function? What can you conclude from this? – 2017-01-24
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    @aduh the way Rudin defines it: if $f=u+iv$ as $\int_Nu^+-\int_Nu^++i\int_Nv^+-i\int_Nv^-$. The only thing I see from this immediately though is that if $\int_N|f|d\mu<\infty$, then each of the the integrals is also $<\infty$ and so $\int_Nfd\mu$ makes sense. – 2017-01-24

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We have

$$\int_\mathbb{N}u^+d\mu = \sum_{n=1}^\infty u(n)^+$$

$$\int_\mathbb{N}u^-d\mu = \sum_{n=1}^\infty u(n)^-$$

$$\int_\mathbb{N}v^+d\mu = \sum_{n=1}^\infty v(n)^+$$

$$\int_\mathbb{N}v^-d\mu = \sum_{n=1}^\infty v(n)^-$$

Where each sum is convergent since as you said each integral is finite. Rudin defines the integral to be

$$\int_\mathbb{N}fd\mu = \int_\mathbb{N}u^+d\mu - \int_\mathbb{N}u^-d\mu + i\int_\mathbb{N}v^+d\mu - i\int_\mathbb{N}v^-d\mu$$

$$=\sum_{n=1}^\infty u(n)^+ - \sum_{n=1}^\infty u(n)^- + i\sum_{n=1}^\infty v(n)^+ - i\sum_{n=1}^\infty v(n)^+$$

Since each sum is convergent, moving the $i$'s inside the sums we can add them term by term.

$$=\sum_{n=1}^\infty\bigg[u(n)^+ - u(n)^- + iv(n)^+ - iv(n)^-\bigg]$$

$$=\sum_{n=1}^\infty f(n)$$