Integral of Integrable Function is Monotone

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.

Suppose that $f \le g$ pointwise.

Then:


 * $\displaystyle \int f \rd \mu \le \int g \rd \mu$

Proof
Since:


 * $f \le g$

we have that $g - f$ is well-defined with:


 * $g - f \ge 0$

From Integral of Integrable Function is Additive: Corollary 2, we have:


 * $g - f$ is $\mu$-integrable

with:


 * $\ds \int \paren {g - f} \rd \mu = \int g \rd \mu - \int f \rd \mu$

Since:


 * $\ds \int \paren {g - f} \rd \mu \ge 0$

we have:


 * $\ds \int g \rd \mu - \int f \rd \mu \ge 0$

This gives:


 * $\ds \int g \rd \mu \ge \int f \rd \mu$