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

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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
$(2) \quad$ $\ds \int \size f \rd \mu = 0$


Corollary

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

Let $f: X \to \overline \R$ be a non-negative integrable function.

Let $A, B \in \Sigma$ have $A \subseteq B$.


Then:

$\ds \int_A f \rd \mu = \int_B f \rd \mu$

if and only if:

$f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.


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:

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

$\Box$


$(2)$ implies $(1)$

Suppose that:

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

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

\(\ds \map \mu {\set {x \in X : \size {\map f x} \ge \frac 1 n} }\) \(\le\) \(\ds n \int \size f \rd \mu\)
\(\ds \) \(=\) \(\ds 0\)

So:

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

for each $n \in \N$.

Note that:

\(\ds \set {x \in X : \map f x \ne 0}\) \(=\) \(\ds \set {x \in X : \size {\map f x} > 0}\)
\(\ds \) \(=\) \(\ds \bigcup_{n \mathop = 1}^\infty \set {x \in X : \size {\map f x} \ge \frac 1 n}\)

From Measure is Countably Subadditive, we have:

\(\ds \map \mu {\set {x \in X : \map f x \ne 0} }\) \(=\) \(\ds \mu \paren {\bigcup_{n \mathop = 1}^\infty \set {x \in X : \size {\map f x} \ge \frac 1 n} }\)
\(\ds \) \(\le\) \(\ds \sum_{n \mathop = 1}^\infty \map \mu {\set {x \in X : \size {\map f x} \ge \frac 1 n} }\)
\(\ds \) \(=\) \(\ds 0\)

So:

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

That is:

$f = 0$ almost everywhere.

$\blacksquare$


Sources

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