# Let $\varphi_n, \psi$ be simple functions such that $(\varphi_n)$ is increasing and $\psi\le\lim_n\varphi_n$. Then $\int\psi\le\lim_n \int \varphi_n$

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 $$\lambda \searrow 1$$.

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 $$\int \psi \leq \int \lim _{n \to \infty}\varphi_n = \lim_{n \to \infty} \int \varphi_n$$.