Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero

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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:

$\ds \int \size f \rd \mu = \int f^+ \rd \mu + \int f^- \rd \mu = 0$

$\blacksquare$

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