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
- $(2) \quad$ $\ds \int \size f \rd \mu = 0$
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:
- $\ds \int_A f \rd \mu = \int_B f \rd \mu$
- $f \times \chi_{B \setminus A} = 0$ $\mu$-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
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $10.9 \ \text{(i)}$