Triangle Inequality for Integrals/Corollary

Theorem

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

Let $f: X \to \overline \R$ be a $\mu$-integrable function be such that:

$\ds \int \size f \rd \mu = 0$

Then:

$\ds \int f \rd \mu = 0$

Proof

From Triangle Inequality for Integrals, we have:

$\ds \size {\int f \rd \mu} \le \int \size f \rd \mu$

We have:

$\ds \int \size f \rd \mu = 0$

so:

$\ds \size {\int f \rd \mu} \le 0$

That is:

$\ds \size {\int f \rd \mu} = 0$

so:

$\ds \int f \rd \mu = 0$

$\blacksquare$