Integral of Characteristic Functions

This is week three of re-writing a proof from Axler’s book Measure, Integration & Real Analysis by Sheldon Axler. The proof covered here demonstrates the closed form for integral of characteristic function.

Def. (\(\mathcal{S}\)-Partion): Suppose \(\mathcal{S}\) is a \(\sigma\)-algebra on a set \(X\). An \(\mathcal{S}\)-partion of \(X\) is a finite collection \(A_1,\dots ,A_m\) of disjoint sets in \(\mathcal{S}\) such that \(A_1 \cup\dots \cup A_m =X\).

Def. (Lower Lebesgue Sum): \(\mathcal{L}(f,P)=\sum_{j=1}^{m}\mu (A_j )\inf_{A_j}f\)

Def. (Integral of a Nonnegative Function): \(\int f d\mu = \sup \{ \mathcal{L}(f,P) : \; P \; \text{is a partion of } X \}=\sup_{A_j\in P} \sum_{j=1}^{m}\mu (A_j )\inf_{A_j} f\)

Lemma 1: Suppose \((X,\mathcal{S},\mu )\) is a measure space and \(D,E\in\mathcal{S}\) are such that \(D\subseteq E\). Then \(\mu (E\setminus D)=\mu (E)-\mu (D)\) provided that \(\mu (D)<\infty\).

Integral of a Characteristic Function: Suppose \((X,\mathcal{S},\mu )\) is a measure space and \(E\in \mathcal{S}\). Then \(\int \chi_{E} d\mu =\mu (E)\).

Proof: \((\int \chi_{E}d\mu \geq \mu (E)):\) say \(P\) is a partion of \(X\) with \(E\) and its compliment \(E^{c}=X\setminus E\). Then \(\mathcal{L}(\chi_{E},\{ E,E^{c} \})=\mu (E)\cdot 1+\mu (E^c)\cdot 0 =\mu (E)\). Yet, whose to say \(P=\{ E,E^c \}\) is the \(\sup_{P}\mathcal{L}(\chi_{E},P)\) hence, \((\int \chi_{E}d\mu \geq = \mathcal{L}(\chi_{E},\{E,E^c\}) = \mu (E))~.\)

\((\int \chi_{E}d\mu \leq \mu (E)):\) say \(P = \{ A_1,\dots ,A_m \}\) partions \(X\). Note, \(\mu (A_j)\inf_{A_j}\chi_{E} = \mu (A_j)\cdot 1\) if \(A_j\subset E\) by Def. \(\chi_{E}\) and \(\mu (A_j)\cdot 0\) otherwise. Thus,

and therefore \(\int \chi_{E}d\mu \leq \mu (E)\). \(\square\)