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Equip $\mathbb{N}$ with counting measure $\#$. For a nonnegative function $f:\mathbb{N}\to [0,\infty ]$, defineThis lies in $[0,\infty ]$ and agrees with the series as a Lebesgue integral against counting measure.

For $f:\mathbb{N}\to \mathbb{C}$, definewhen $f$ is integrable, which here meansThus the complex sum is defined under absolute convergence and can be obtained by integrating the real and imaginary parts.

A simple function on $\mathbb{N}$ is a function with finite range, equivalentlyfor a finite measurable partition by subsets $A_j\subseteq \mathbb{N}$. Its nonnegative sum is ${\sum }_j{a}_j\#A_j$. It is finite exactly when every level set with $a_j>0$ is finite, equivalently when $s$ has finite support after its zero level is discarded.

The Beppo Levi theorem, or monotone convergence theorem, states that if $(f_n)$ is a sequence of nonnegative measurable functions on a measure space andpointwise almost everywhere, then $f$ is measurable andwhere either side may be infinite.

Measurability of $f={\sup }_n{f}_n$ follows from measurability of countable suprema. Since $f_n\le f$, monotonicity of the Lebesgue integral givesFor the reverse inequality, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. DefineThen $E_n\uparrow X$ up to the null set on which convergence fails. Since $f_n\ge c,s{1}_{E_n}$,by continuity from below of a measure, applied to the finitely many level sets of $s$. Hence ${\lim }_n\int f_n\ge c\int s$. Letting $c\uparrow 1$ and taking the supremum over all simple $s\le f$, as in the definition of the Lebesgue integral, gives the reverse inequality and proves the theorem.