Characterization of Almost Everywhere Zero

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

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

Let $f : X \to \overline \R$ be a measurable function.

Then:
 * $\ds \forall A \in \Sigma : \int \paren {\chi_A \cdot f} \rd \mu = 0$


 * $f = 0$ $\mu$-almost everywhere
 * $f = 0$ $\mu$-almost everywhere

where:
 * $\chi_A$ is the characteristic function of $A$

Necessary condition
First, by Measurable Function Zero A.E. iff Absolute Value has Zero Integral:
 * $\ds \int \size f \rd \mu = 0$

Let $A\in\Sigma$.

Let $\paren {\chi_A \cdot f}^+$, $\paren {\chi_A \cdot f}^-$ be the positive and negative parts of $\chi_A \cdot f$, respectively.

Observe:
 * $\paren {\chi_A \cdot f}^+ \le \size f$

and:
 * $\paren {\chi_A \cdot f}^- \le \size f$

Therefore, by Integral of Positive Measurable Function is Monotone:
 * $\ds \int \paren {\chi_A \cdot f}^+ \rd \mu \le \int \size f \rd \mu = 0$

and:
 * $\ds \int \paren {\chi_A \cdot f}^- \rd \mu \le \int \size f \rd \mu = 0$

This means by definition of integral:
 * $\ds \int\chi_A \cdot f \rd \mu = 0$.

Sufficient condition
Similarly:
 * $\map \mu {\set {-f > 1 / n} } = 0$

Altogether:
 * $\map \mu {\set {\size f > 1 / n} } = 0$

Therefore: