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:

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.