Counting measure

Take the measure space (\mathbb{N}, 2^\mathbb{N}, \mu), where 2^\mathbb{N} is the set of all subsets of the naturals and \mu the counting measure. Take any measurable f : \mathbb{N} \to [0,\infty]. As it is defined on \mathbb{N}, f can be represented pointwise as f(x) = \sum_{n=1}^\infty f(n) 1_{\{n\}}(x) = \lim_{M \to \infty} \underbrace{ \ \sum_{n=1}^M f(n) 1_{\{n\}}(x) \ }_{ \phi_M (x) } = \lim_{M \to \infty} \phi_M (x) Each \phi_M is measurable. Moreover \phi_{M+1}(x) = \phi_M (x) + f(M+1) \cdot 1_{ \{M+1\} }(x) \geq \phi_M (x) . Still further, as each \phi_M is a simple function \int_\mathbb{N} \phi_M d\mu = \int_\mathbb{N} \left( \sum_{n=1}^M f(n) 1_{\{n\}} (x) \right) d\mu = \sum_{n=1}^M f(n) \mu (\{n\}) = \sum_{n=1}^M f(n) \cdot 1 = \sum_{n=1}^M f(n) Hence by the monotone convergence theorem \int_\mathbb{N} f d\mu = \lim_{M \to \infty} \int_\mathbb{N} \phi_M d\mu = \lim_{M \to \infty} \sum_{n=1}^M f(n) = \sum_{n=1}^\infty f(n) ==Discussion==

The counting measure is a special case of a more general construction. With the notation as above, any function f : X \to [0, \infty) defines a measure \mu on (X, \Sigma) via \mu(A):=\sum_{a \in A} f(a)\quad \text{ for all } A \subseteq X, where the possibly uncountable sum of real numbers is defined to be the supremum of the sums over all finite subsets, that is, \sum_{y\,\in\,Y\!\ \subseteq\,\mathbb R} y\ :=\ \sup_{F \subseteq Y,\, |F| Taking f(x) = 1 for all x \in X gives the counting measure. ==See also==