Measurable Function Zero A.E. iff Absolute Value has Zero Integral

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

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

Then the following are equivalent:


 * $(1) \quad$ $f = 0$ almost everywhere
 * $(2) \quad$ $\ds \int \size f \rd \mu = 0$

Proof
Let $\EE^+$ be the space of positive simple functions.

$(1)$ implies $(2)$
Suppose that:


 * $f = 0$ almost everywhere.

Note that if $\map f x = 0$ for some $x \in X$, then:


 * $\size {\map f x} = 0$ for some $x \in X$.

So:


 * $\size f = 0$ almost everywhere.

That is:


 * there exists a null set $N \subseteq X$ such that if $\size {\map f x} \ne 0$, then $x \in N$.

From Absolute Value of Measurable Function is Measurable:


 * $\size f$ is $\Sigma$-measurable.

So, its $\mu$-integral is well-defined.

Let $g \in \EE^+$ have $g \le \size f$.

Then, if $\map g x \ne 0$ for $x \in X$ we must have $\size {\map f x} \ne 0$.

So, if $x \in X$ has $\map g x \ne 0$, then $x \in N$.

From Simple Function has Standard Representation:


 * there exists disjoint $\Sigma$-measurable sets $E_1, E_2, \ldots, E_n$ and non-negative real numbers $a_1, a_2, \ldots, a_n$ such that:


 * $\ds \map f x = \sum_{i \mathop = 1}^n a_i \map {\chi_{E_i} } x$


 * for each $x \in X$.

If $a_i \ne 0$, then:


 * $\map f x \ne 0$ for $x \in E_i$.

That is:


 * $x \in N$ for all $x \in E_i$.

So, we obtain:


 * $E_i \subseteq N$

for each $i$.

Then, from Measure is Monotone, we have:


 * $\map \mu {E_i} \le \map \mu N = 0$

So:


 * $\map \mu {E_i} = 0$

for each $i$ such that $a_i \ne 0$.

So:


 * $a_i \map \mu {E_i} = 0$

for each $i$.

Then:


 * $\ds \map {I_\mu} g = 0$

for all $g \in \EE^+$ with $g \le \size f$, where:


 * $\map {I_\mu} g$ denotes the $\mu$-integral of the positive simple function $g$.

So, from the definition of the $\mu$-integral:


 * $\ds 0 = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+} = \int \size f \rd \mu$

$(2)$ implies $(1)$
Suppose that:


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

From Markov's Inequality, we have, for each $n \in \N$:

So:


 * $\ds \map \mu {\set {x \in X : \size {\map f x} \ge \frac 1 n} } = 0$

for each $n \in \N$.

Note that:

From Measure is Countably Subadditive, we have:

So:


 * $\map \mu {\set {x \in X : \map f x \ne 0} } = 0$

That is:


 * $f = 0$ almost everywhere.