[Math] Lebesgue integral with repect to counting measure

measure-theory

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:

  1. $A_j^n$ 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 $F_n := \bigcup_{j=1}^{m_n} A_j^n$, $$\int f_n \, d\mu = \sum_{j=1}^{m_n} \sum_{x \in A_j^n} c_j^n \leq \sum_{x \in F_n} f(x) \leq \sup\left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\}.$$
  2. 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,

$$\int f \, d\mu \leq \sup \left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\}.$$

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