Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.
Let $f : X \to \overline \R$ be a $\GG$-integrable function.
Suppose that, for all $G \in \GG$:
- $\ds \int_G f \rd \mu = 0$
Then $f = 0$ $\mu$-almost everywhere.
Proof
Let:
- $\ds A = \set {x \in X : \map f x \ge 0}$
so that:
- $\ds X \setminus A = \set {x \in X : \map f x < 0}$
From Characterization of Measurable Functions:
- $A$ and $X \setminus A$ are $\GG$-measurable.
So:
- $\ds \int_A f \rd \mu = 0$
That is:
- $\ds \int f \cdot \chi_A \rd \mu = 0$
Note that for $x \in A$, we have:
- $\map f x \map {\chi_A} x = \map f x \ge 0$
and for $x \in X \setminus A$, we have:
- $\map f x \map {\chi_A} x = 0$
So:
- $f \cdot \chi_A \ge 0$
So that:
- $f \cdot \chi_A = \size {f \cdot \chi_A}$
This gives:
- $\ds \int \size {f \cdot \chi_A} = 0$
Then, from Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have:
- $f \cdot \chi_A = 0$ $\mu$-almost everywhere.
That is, there exists a $\mu$-null set $N_1 \subseteq X$ such that:
- if $\map f x \map {\chi_A} x \ne 0$ then $x \in N_1$.
Note that we also have:
- $\ds \int_{X \setminus A} f \rd \mu = 0$
so that:
- $\ds \int f \cdot \chi_{X \setminus A} \rd \mu = 0$
From Integral of Integrable Function is Homogeneous, we have:
- $\ds \int \paren {-f \cdot \chi_{X \setminus A} } \rd \mu = 0$
Note that for $x \in A$, we have:
- $-\map f x \map {\chi_{X \setminus A} } x = 0$
and for $x \in X \setminus A$, we have:
- $-\map f x \map {\chi_{X \setminus A} } x = -\map f x > 0$
So:
- $-f \cdot \chi_{X \setminus A} \ge 0$
so that:
- $-f \cdot \chi_{X \setminus A} = \size {f \cdot \chi_{X \setminus A} }$
This gives:
- $\ds \int \size {f \cdot \chi_{X \setminus A} } \rd \mu = 0$
Then, from Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have:
- $f \cdot \chi_{X \setminus A} = 0$ $\mu$-almost everywhere.
That is, there exists a $\mu$-null set $N_2 \subseteq X$ such that:
- if $\map f x \map {\chi_{X \setminus A} } x \ne 0$ then $x \in N_2$.
From Characteristic Function of Set Difference, we have:
- $\chi_{X \setminus A} = \chi_X - \chi_A$
So:
\(\ds f \cdot \chi_{X \setminus A} + f \cdot \chi_A\) | \(=\) | \(\ds f \cdot \paren {\chi_{X \setminus A} + \chi_A}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds f \cdot \chi_X\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds f\) | since $f$ is a function $X \to \overline \R$ |
So if $x \in X$ has $\map f x \ne 0$ then:
- either $\map f x \map {\chi_{X \setminus A} } x \ne 0$ or $\map f x \map {\chi_A} x \ne 0$.
That is:
- either $x \in N_1$ or $x \in N_2$.
So if $\map f x \ne 0$, we have that:
- $x \in N_1 \cup N_2$
From Null Sets Closed under Countable Union, we have that:
- $N_1 \cup N_2$ is a $\mu$-null set.
So:
- $f = 0$ $\mu$-almost everywhere.
$\blacksquare$
Proof 2
In view of Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we shall show:
- $\ds \int \size f \rd \mu = 0$
Since $f$ is $\GG$-measurable:
- $G_+ := \set { x \in X : f(x) > 0 } \in \GG$
and:
- $G_- := \set { x \in X : f(x) \le 0 } \in \GG$
On the one hand:
\(\ds \int f^+ \rd \mu\) | \(=\) | \(\ds \int f \cdot \chi _{ G_+ } \rd \mu\) | $f^+$ is positive part of $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int _{ G_+ } f \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | by hypothesis |
On the other hand, from $f \cdot \chi _{ G_- } \le 0$ follows:
\(\ds \paren {f \cdot \chi _{ G_- } } ^+\) | \(=\) | \(\ds 0\) |
and:
\(\ds \paren {f \cdot \chi _{ G_- } } ^-\) | \(=\) | \(\ds - \paren {f \cdot \chi _{ G_- } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren { -f } \cdot \chi _{ G_- }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds f^-\) | $f^-$ is negative part of $f$ |
Thus:
\(\ds \int _{ G_- } f \rd \mu\) | \(=\) | \(\ds \int f \cdot \chi _{ G_- } \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {f \cdot \chi _{ G_- } } ^+ \rd \mu - \int \paren {f \cdot \chi _{ G_- } } ^- \rd \mu\) | integral of $f \cdot \chi _{ G_- }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds - \int f ^- \rd \mu\) |
which implies:
\(\ds \int f ^- \rd \mu\) | \(=\) | \(\ds - \int f \cdot \chi _{ G_- } \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds - \int _{ G_- } f \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | by hypothesis |
Therefore:
- $\ds \int \size f \rd \mu = \int f^+ \rd \mu + \int f^- \rd \mu = 0$
$\blacksquare$