Lebesgue integral with repect to counting measure Let $\Omega$ be a set, $\mathcal{A} = \mathcal{P}(\Omega)$ the power set, $\mu$ counting measure, $f$ a nonnegative function on $\Omega$, I want to show that $$ \int_\Omega f d\mu = \sum_{x \in \Omega} f(x) $$ where $\sum_{x \in \Omega} f(x) = \sup \{\sum_{x \in F} f(x): F \mbox{ finite },\, F \subset \Omega \}$.
$\forall F \subset \Omega$, we have
$$ \int_{\Omega} f d\mu \ge \int_F f d\mu = \sum_{x \in F} \int_{\{x\}} f d\mu = \sum_{x \in F} f(x) \mu(\{x\}) = \sum_{x \in F} f(x)$$
so $ \int_{\Omega} f d\mu \ge \sup_{F \subset \Omega} \{ \sum_{x \in F} f(x) \}$. How do I show the reverse inequality ?
 A: As $\mathcal{A} = \mathcal{P}(\Omega)$, we know that any function $f: \Omega \to \mathbb{R}$ is measurable. By the Sombrero lemma, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of non-negative simple functions such that $f_n \uparrow f$. Applying Beppo Lévy's theorem yields
$$\int f \, d\mu = \sup_{n \in \mathbb{N}} \int f_n \, d\mu. \tag{1}$$
As $f_n$ is a non-negative simple function, we can choose $c_j^n > 0$ and $\emptyset \neq A_j^n \in \mathcal{B}(\mathbb{R}) $ such that
$$f_n(x) = \sum_{j=1}^{m_n} c_j^n \cdot 1_{A_j^n}(x).$$
Now we consider two cases separately:


*

*$A_j^n$ is a finite set for all $n \in \mathbb{N}$, $j=1,\ldots,m_n$. Then, as $\mu$ is the counting measure, we have $$\int f_n \, d\mu = \sum_{j=1}^{m_n} \sum_{x \in A_j^n} c_j^n.$$ As $f_n \leq f$, we know that $c_j^n \leq f(x)$ for any $x \in A_j^n$. Hence, if we set $F_n := \bigcup_{j=1}^{m_n} A_j^n$, $$\int f_n \, d\mu = \sum_{j=1}^{m_n} \sum_{x \in A_j^n} c_j^n \leq \sum_{x \in F_n} f(x) \leq \sup\left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\}.$$

*There exists $n \in \mathbb{N}$ and $j\in \{1,\ldots,m_n\}$ such that $A_j^n$ is infinite. Then $$\int f \, d\mu \geq \int f_n \, d\mu \geq \int_{A_j^n} f_n \, d\mu = \infty.$$ On the other hand, we can choose finite sets $F_m$ such that $F_m \subseteq F_{m+1} \subseteq A_j^n$ and $\bigcup_m F_m$ is an infinite set, then $$\begin{align*} \sup \left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\} &\geq \lim_{m \to \infty} \sum_{x \in F_m} f(x) \\ &\geq \lim_{n \to \infty} \sum_{x \in F_m} \underbrace{f_n(x)}_{c_j^n} \\ &= c_j^n \lim_{m \to \infty} \sum_{x \in F_m} 1 = \infty. \end{align*}$$


Consequently, in both cases,
$$\int f \, d\mu \leq \sup \left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\}.$$
A: I only have a partial answer, assuming that $\int_\Omega f\; d\mu<\infty$. Call $\alpha$ the supremum in the question and let $E_n=\{x\in\Omega: f(x)>\frac{1}{n}\}$. Since $f$ is integrable, for every positive integer $n$, $\frac{1}{n}\mu(E_n)=\int_\omega \frac{1}{n}\chi_{E_n}\; d\mu\leq \int_\Omega f; d\mu<\infty$ so $\mu(E_n)<\infty$ for every $n$. Because $\mu$ is the counting measure, we have that $E_n$ is finite for every $n$. Since $E_n\subset E_{n+1}$ and the mapping $A\mapsto \int_A f\; d\mu$ defined on $\mathcal{P}(\Omega)$ is a measure, it follows, by properties of the measure, that
$$
\lim\limits_{n\to\infty}\int_{E_n} f\; d\mu=\int_E f\; d\mu
$$
Where $E=\{x\in\Omega: f(x)>0\}$. Note that $f$ is zero on the complement of $E$ so $\int_E f\; d\mu=\int_\Omega f\; d\mu$.
Given $\varepsilon>0$, let $N$ be large enough to guarantee that 
$$
\int_\Omega f\; d\mu -\int_{E_N}f\; d\mu<\varepsilon
$$
Since $E_N$ is finite, the last integral equals to $\sum_{x\in E_N}f(x)$. Hence
$$
\int_\Omega f\; d\mu<\sum_{x\in E_N}f(x)+\varepsilon\leq \alpha+\varepsilon
$$
Since $\varepsilon$ was arbitrary, if follows that $\int_\Omega f\; d\mu\leq \alpha$ as desired.
