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:


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