Triangle Inequality for Integrals/Corollary
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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:
- $\ds \int f \rd \mu = 0$
$\blacksquare$