Let $u: \mathbb{N} \to \mathbb{R}^+$ be a positive sequence. Then it is true that
$$
\int u d\mu = \sum_{n=1}^\infty u(n).
$$
For the measure space $(\mathbb{N}, \mathcal{P}(\mathbb{N}),\mu)$, where $\mu$ is the counting measure. The proof uses monotone convergence theorem, and goes as follows:
Take the measure space $(\mathbb{N},\mathcal{P}(\mathbb{N}), \underbrace{\sum_{j=1}^\infty \delta_j}_{\mu})$.
Let $u: \mathbb{N} \to \mathbb{R}^+$ be a positive function ($u \geq 0$).
We define
$$
u_n(k) =
\begin{cases}
u(k) & k \leq n, \\
0 & k > n.
\end{cases}
$$
For all $n \in \mathbb{N}$, $u_n$ is a normal function since we have $u_n = \sum_{k=1}^n u(k)1_{\{k\}}$. Thus, the approximation cuts of $u$ after $n$. Moreover, we have
$$
\int u_n d\mu = I_\mu(f) = \sum_{k=1}^n u(k) \mu(\{k\}) = \sum_{k=1}^n u(k).
$$
Since we have $0 \leq u_n \uparrow u$, then it follows by MCT that
$$
\int_{\mathbb{N}} u d\mu \stackrel{MTC}{=} \lim_{n \to \infty} \int u_n d\mu = \sum_{k=1}^\infty u(k).
$$
Now can someone tell me under what condition this is true if $u: \mathbb{N} \to \mathbb{R}$ is not necessarily positive?
Best Answer
I prove
By the definition of integral we have $\int u d\mu = \int u^+ d\mu - \int u^- d\mu$
By what you just showed for $u^+,u^-$ we have $u^\pm = \sum_n u^\pm (n)$
Since the sum is uniformly convergence it is the same to sum from $1,2,...$ or to sum first the $n$ for which $u(n)>0$ and only then the $n$ for which $u(n)<0$.