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 ?
Best Answer
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:
Consequently, in both cases,
$$\int f \, d\mu \leq \sup \left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\}.$$