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Let $(X, \mathcal A, \mu)$ be a complete measure space and $E := [0, \infty]$ endowed with order topology.

  • $f \in E^{X}$ is called $\mu$-simple if $f = \sum_{k=1}^n e_k 1_{A_k}$ where $e_k \in E \setminus \{\infty, 0\}$ and $(A_k)_{k=1}^n$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. The Bochner integral of such $f$ w.r.t. $\mu$ is defined by $\int f := \sum_{k=1}^n e_k \mu(A_k)$. Let $\mathcal S (X, \mu, E)$ be the space of $\mu$-simple functions.

  • $f \in E^{X}$ is called $\mu$-measurable if $f$ is a $\mu$-a.e. limit of a sequence $(f_n) \subset \mathcal S (X, \mu, E)$. Let $\mathcal L_0 (X, \mu, E)$ be the space of such $\mu$-measurable functions.

I'm reading below result from Amann's Analysis III.

Lemma 1.3: Let $\varphi_n, \psi \in \mathcal S (X, \mu, E)$ such that the sequence $(\varphi_n)$ is increasing and $\psi \le \lim_n \varphi_n$. Then $$ \int \psi \le \lim_n \int \varphi_n. $$

Below proof (by Amann) is elegant and short due to the clever introduction of $B_n$ and $\lambda$.

My question: Are there other approaches that are easier (or more natural) to come up with?


Proof: Fix $\lambda>1$ and let $B_n := \{x \in X | \lambda \varphi_n (x) \ge \psi (x)\}$.

It follows from $(X, \mathcal A, \mu)$ is complete that $B_n \in \mathcal A$. Because $(\varphi_n)$ is increasing, $B_n \subset B_{n+1}$. Since $\psi \le \lim_n \varphi_n$ and $\lambda > 1$, we have $X =\cup_n B_n$. It follows that $(A\cap B_n) \nearrow A$ for any $A \in \mathcal A$. By continuity of measure from below, we have $$ \mu(A) = \lim_n \mu(A \cap B_n) \quad \forall A \in \mathcal A. $$

Assume $\psi = \sum_{i=1}^p e_i 1_{A_i}$ where $e_i \in E \setminus \{\infty, 0\}$ and $(A_i)_{i=1}^p$ is a finite sequence of pairwise disjoint sets with finite measure in $\mathcal A$. Then $$ \begin{align} \int \psi &= \sum_{i=1}^p e_i \mu(A_i) \\ &= \sum_{i=1}^p e_i \lim_n \mu(A_i \cap B_n) \\ &\overset{(\star)}{=} \lim_n \sum_{i=1}^p e_i \mu(A_i \cap B_n) \\ &= \lim_n \int \psi1_{B_n} \\ &\le \lim_n \int \lambda \varphi_n \\ &=\lambda \lim_n \int \varphi_n. \end{align} $$

The interchange of $\lim_n$ and $\sum_{i=1}^p$ in $(\star)$ is valid because the sum is finite. The claim then follows by taking the limit $\lambda \searrow 1$.

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If you know the monotone convergence theorem, this is simple:

By the monotone convergence theorem, we have $\lim_{n \to \infty} \int \varphi_n = \int \lim _{n \to \infty}\varphi_n$. Since $\psi \leq \lim_{n \to \infty}\varphi_n$, monotonicity of the integral yields $\int \psi \leq \int \lim _{n \to \infty}\varphi_n = \lim_{n \to \infty} \int \varphi_n$.

The monotone convergence theorem can be proved using an argument resembling what you presented.

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  • $\begingroup$ Because Lemma 1.3 is introduced even before the definition if integral. I could not use MCT. $\endgroup$
    – Akira
    Commented Aug 10, 2022 at 23:17

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