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:

$\ds \int f \rd \mu \le \int g \rd \mu$

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:

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

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $10.4 \ \text{(iv)}$