Monotone Convergence Theorem

This is week four of re-writing a proof from the book Measure, Integration & Real Analysis by Sheldon Axler. The proof covered here demonstrates the Monotone Convergence Theorem.

Lemma 1: \(\int f d\mu = \sup \{ \sum_{j=1}^{m} \mu (A_j ):A_1,\dots ,A_m \text{ are disjoint sets in } \mathcal{S}, c_1,\dots ,c_m \in [0,\infty ], f(x)\geq \sum_{j=1}^{m}c_j \chi_{A_j} \; \forall \; x\in X \}\)

Lemma 2: Suppose \((X,\mathcal{S},\mu )\) is a measure space and \(E_1\subseteq E_2\subseteq \dots\) is an increasing sequence in \(\mathcal{S}\), then \(\mu (\cup_{k=1}^{\infty} E_k ) = \lim_{k\to \infty}\mu (E_k)\).

Monotone Convergence Theorem: Suppose \((X,\mathcal{S},\mu )\) is a measure space and \(0\leq f_1 \leq f_2 \leq \dots\) is an increasing sequence of \(\mathcal{S}\)-measurable functions. Define \(f:X\to [0,1]\) by \(f(x)=\lim_{k\to \infty} f_k (x)\). Then \(\lim_{k\to \infty}\int f_k d\mu = \int fd\mu\).

Proof: \((\lim_{k\to \infty}\int f_k d\mu \leq \int f d\mu ):\) note that since each \(f_i\) is \(\mathcal{S}\)-measurable its holds that \(f\) is \(\mathcal{S}\)-measurable. Furtheremore, since \(f_k(x)\leq f(x)\) for every \(x\in X\), we have that \(\int f_k(x)d\mu \leq \int fd\mu\) for each \(k\in \mathbb{Z}_{+}\). Hence, we have that \(\lim_{k\to \infty}\int f_k d\mu \leq \int f d\mu\).

\((\lim_{k\to \infty}\int f_k d\mu \geq \int f d\mu ):\) suppose \(A_1,\dots ,A_m\) are disjoint sets in \(\mathcal{S}\) and \(c_1,\dots ,c_m \in [0,\infty ]\) such that \(f(x)\geq \sum_{j=1}^{m}c_j\chi_{A_j}(x)\) for every \(x\in X\). Essentiaally, the choosen set \(A_1,\dots ,A_m\) are used to define a simple function which lower bounds \(f\). Let \(t\in (0,1)\). For \(k\in \mathbb{Z}_{+}\), let \(E_k =\{ x\in X\mid f_k(x)\geq t\sum_{j=1}^{m} c_j \chi_{A_j}(x) \}\) (Eq. 1). By construction, \(E_1\subseteq E_2\subseteq \dots\) is an increasing sequence of sets in \(\mathcal{S}\) whose union equals \(X\). Note, Lemma 2 applies and we have \(\lim_{k\to \infty}\mu (A_j\cap E_k)=\mu (A_j)\) for each \(j\in \{1,\dots ,m \}\). Thus, by Lemma 3 we have that \(\int f_k d\mu \leq t\sum_{j=1}^{m}c_j \mu (A_j\cap E_k)\). By taking the lmit as \(k\to \infty\) and \(t\to 1\), we get \(\lim_{k\to \infty}\int f_kd\mu \geq \sum_{j=1}^{m}c_j \mu (A_j)\). As \(A_1,\dots ,A_m\) was arbitrarly declared, let \(A_1,\dots ,A_m\) be the sup \(\mathcal{S}\)-partion of \(X\) and all \(c_1,\dots c_m\in [0,\infty]\) satisfying Eq 1. Finally, by Lemma 1, we get \(\lim_{k\to \infty}\int f_k d\mu \geq \int f d\mu\) completing the proof. \(\square\)