TITLE: Fatou's Lemma and Counting Measure
QUESTION [10 upvotes]: I have a vague problem in a Measure and Integration book here. They ask me to consider $\mu$ the counting measure in $\mathbb{N}$ and interpret Fatou's lemma, monotone and dominated convergence theorems as statements about infinite series. I thought it would be easy but got stuck really quick...
If we consider a sequence of non negative functions $f_n\in L^1$, Fatou's lemma says that $$\int_\mathbb{N}\liminf f_n \ d\mu \leq \liminf\int_\mathbb{N}f_n \ d\mu.$$
The real problem comes when I try to understand what are this integrals. Starting with a simple function $\Phi =\sum_{i=1}^kx_i\mathcal{X}_{E_i}$, we have that $\int_\mathbb{N}\Phi \ d\mu = \sum_{i=1}^kx_i\mu(E_i) = \sum_{i=1}^kx_i|E_i|$, where $|E_i|$ is the number of elements in $E_i$. Each function $f_n$ is the $\sup$ of simple functions, but it's not clear how I should use all this together. Even if I consider a sequence $0\leq \Phi_1\leq\Phi_2\leq\ldots\leq f_n$ converging to $f_n$, they are not partial sums. 
To make things worse, there is infinite $f_n$ to consider and the $\liminf$ after that. Any help is welcome to interpret all this.
Thank you.

REPLY [2 votes]: I would like to provide a clearer proof to the post. As a start, we recall Fatou's lemma, that is, $$\int_{\Omega}\liminf\:h \:d\mu \leq \liminf \int_{\Omega}h\:d\mu $$
We first show the sketch of the proof:
$\textbf{First}$, we show that for any $h$: $(\mathbb{N},P(\mathbb{N})) \rightarrow (\mathbb{R}^{+},B(\mathbb{R}^+))$ (i.e. nonnegative measurable function), we have
\begin{equation}
\sum_{k=1}^{\infty}h(k)=\int_{\mathbb{N}}h(k)\:d\mu_{c}(k), \tag{1}
\end{equation}
where $\mu_{c}$ is a counting measure, $\mu_c$ is defined on $(\mathbb{N},P(\mathbb{N}))$, where $P(\mathbb{N})$ is the powerset of natural numbers.
$\textbf{Second}$, let us define $f(k) \equiv \liminf_{n\rightarrow \infty}\:f_n(k) \;\forall k\in\mathbb{N} $. Since this newly defined $f$ is such that $f$: $\mathbb{N} \rightarrow \mathbb{R}^+$ and is measurable, then from (1), $$\sum_{k=1}^{\infty} \liminf_{n\rightarrow\infty}f_n(k) = \int_{\mathbb{N}}\liminf f_n(k)\;d\mu_c \leq \liminf \int_{\mathbb{N}}f_n \;d\mu_c = \liminf \sum_k f_n (k),$$ where the inequality follows from fatou's lemma above, and the last equality follows from (1) above since $f_n$: $\mathbb{N} \rightarrow \mathbb{R}^+$ and is measurable. $\textbf{Our job is therefore reduced to proving (1)}$
$\textbf{Proof of (1)}$:
For any nonnegative measurable function $h$: $(\mathbb{N},P(\mathbb{N})) \rightarrow (\mathbb{R}^{+},B(\mathbb{R}^+))$, we define $h_n \equiv \chi_{[1,n]} h$. Note first that $h_n \rightarrow h$ as $n\rightarrow \infty$. In addition, $h_n$ forms a sequence of nonnegative increasing function.
\begin{align}
\int_{\mathbb{N}} h_n \; d\mu_c =
 \int_{[1,n]} h\; d\mu_c 
&= \int_1 h\; d\mu_c + \int_2 h\; d\mu_c + ... + \int_n h\;d\mu_c \\
&= h(1)+h(2)+...+h(n)=\sum_{k=1}^{n} h(k)\\
\end{align}
Then $\lim_{n\rightarrow\infty} \int_{\mathbb{N}}h_n\;d\mu_c = \lim_{n\rightarrow\infty} \sum_{k=1}^{n} h(k) = \sum_{k=1}^{\infty} h(k) \tag{2}$
However, by $\textbf{monotone convergence theorem}$, we know that $$\lim_{n\rightarrow\infty} \int_{\mathbb{N}}h_n\;d\mu_c = \int_{\mathbb{N}} h\;d\mu_c\tag{3},$$ since $0\leq h_n \uparrow h$. Equating (2) and (3) yields (1)
$ \square$