Prove that $\int$fd$\mu$ $\leq$ $\int$gd$\mu$

Question

Let f, g $\in \mathcal{L}^{+}$ such that f(x) $\leq$ g(x) for all x $\in$ X. Prove that $\int$fd$\mu$ $\leq$$\int$gd$\mu$.

I think I figured out the proof and I am curious as to whether or not I'm on the right track.

Let $\int$fd$\mu$ be a simple function $\phi$

Let $\int$gd$\mu$ be a simple function $\psi$

Using a proposition from Folland's book we have $$\int f d\mu = \int \phi d\mu ~and \int g d\mu = \int \psi d\mu$$

Let E$_{j}$ = $\cup^{m}_{k=1}$(E$_{j} \cap F_{k}$)

Let F$_{k}$ = $\cup^{n}_{j=1}$(E$_{j} \cap F_{k}$) with both being disjoint then we have $$\int \phi = \sum a_{j}\mu(E_{j}\cap F_{k}) \leq \sum b_{k}\mu(E_{j}\cap F_{k}) = \int \psi$$ $$\int fd\mu = \sum a_{j}\mu(E_{j}\cap F_{k}) \leq \sum b_{k}\mu(E_{j}\cap F_{k}) = \int gd\mu$$ $$\int f d\mu \leq g d\mu$$ Whenever E$_{j} \cap F_{k}$ $\neq$ $\oslash$

I was wondering if the proof is correct or if I'm missing something. Thanks again.

Answer 1

A simple proof is this :

First, remember the definition of $\int f d\mu$ :

$$\int f d\mu = \sup \left\{ \int \phi d \mu \mid \phi \leq f \text{ and } \phi \text{ is a simple function} \right\}$$

But it's clear that if $g \leq f$, we have that

$$\left\{ \phi \mid \phi \leq g \text{ and } \phi \text{ is a simple function} \right\} \subset \left\{ \phi \mid \phi \leq f \text{ and } \phi \text{ is a simple function} \right\}$$

So it follow immediatly that

$$\sup \left\{ \int \phi d\mu \mid \phi \leq g \text{ and } \phi \text{ is a simple function} \right\} \\\leq \sup \left\{ \int \phi d\mu \mid \phi \leq f \text{ and } \phi \text{ is a simple function} \right\}$$

Hence

$$\int g d\mu \leq \int f d\mu$$