Integrable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal

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

Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.

Let $f, g: X \to \overline \R$ be $\GG$-integrable functions.

Suppose that, for all $G \in \GG$:


 * $\ds \int_G f \rd \mu = \int_G g \rd \mu$

Then $f = g$ $\mu$-almost everywhere.

Proof
Define:


 * $A = \set {x \in X : \map f x \in \R}$

and:


 * $B = \set {x \in X : \map g x \in \R}$

From Set of Points for which Measurable Function is Real-Valued is Measurable, we have that:


 * $A$ and $B$ are $\GG$-measurable.

From Sigma-Algebra Closed under Countable Intersection, we have that:


 * $A \cap B$ is $\GG$-measurable.

Consider the characteristic function $\chi_{A \cap B}$.

From Characteristic Function Measurable iff Set Measurable, we have that:


 * $\chi_{A \cap B}$ is $\GG$-measurable.

From De Morgan's Laws: Complement of Intersection, we have


 * $X \setminus \paren {A \cap B} = \paren {X \setminus A} \cup \paren {X \setminus B}$

We have that:


 * $X \setminus A = \set {x \in X : \size {\map f x} = +\infty}$

and:


 * $X \setminus B = \set {x \in X : \size {\map f x} = +\infty}$

From Integrable Function is A.E. Real-Valued, we have that:


 * $X \setminus A$ and $X \setminus B$ are $\mu$-null.

From Null Sets Closed under Countable Union, we then have that:


 * $\paren {X \setminus A} \cup \paren {X \setminus B}$ is $\mu$-null.

Let $G \in \GG$.

Note that if $x \in X$ is such that:


 * $\map f x \map {\chi_G} x \map {\chi_{A \cap B} } x \ne \map f x \map {\chi_G} x$

Then:


 * $\map {\chi_{A \cap B} } x = 0$

so:


 * $x \in X \setminus \paren {A \cap B}$

so:


 * $x \in \paren {X \setminus A} \cup \paren {X \setminus B}$

Giving:


 * $f \cdot \chi_G \cdot \chi_{A \cap B} = f \cdot \chi_G$ $\mu$-almost everywhere.

Swapping $f$ for $g$, we also obtain:


 * $g \cdot \chi_G \cdot \chi_{A \cap B} = g \cdot \chi_G$ $\mu$-almost everywhere.

Additionally, from Pointwise Product of Measurable Functions is Measurable, we have that:


 * $f \cdot \chi_G \cdot \chi_{A \cap B}$ and $g \cdot \chi_G \cdot \chi_{A \cap B}$ are $\GG$-measurable for each $G \in \GG$.

By hypothesis, we have that:


 * $\ds \int_G f \rd \mu = \int_G g \rd \mu$

for each $G \in \GG$, so:


 * $\ds \int f \cdot \chi_G \rd \mu = \int g \cdot \chi_G \rd \mu$

Then from A.E. Equal Positive Measurable Functions have Equal Integrals, we have:


 * $\ds \int f \cdot \chi_G \cdot \chi_{A \cap B} \rd \mu = \int g \cdot \chi_G \cdot \chi_{A \cap B} \rd \mu$

so:


 * $\ds \int_G f \cdot \chi_{A \cap B} \rd \mu = \int_G g \cdot \chi_{A \cap B} \rd \mu$

for each $G \in \GG$.

That is:


 * $\ds \int_G \paren {f \cdot \chi_{A \cap B} - g \cdot \chi_{A \cap B} } \rd \mu = 0$

for each $G \in \GG$.

Then, from Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero, we have that:


 * $f \cdot \chi_{A \cap B} - g \cdot \chi_{A \cap B} = 0$ $\mu$-almost everywhere.

That is:


 * $f \cdot \chi_{A \cap B} = g \cdot \chi_{A \cap B}$ $\mu$-almost everywhere.

So, if $x \in X$ is such that $\map f x \ne \map g x$, we must have:


 * $\map {\chi_{A \cap B} } x = 0$

That is:


 * $x \in X \setminus \paren {A \cap B}$

and:


 * $X \setminus \paren {A \cap B}$ is a $\mu$-null set.

So:


 * $f = g$ $\mu$-almost everywhere.

Note
It may be tempting to write that:


 * $\ds \int_G \paren {f - g} \rd \mu = 0$

for each $G \in \GG$, and conclude immediately from Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero.

However, the pointwise difference $f - g$ may not be well-defined.

For example, if there is a point $x \in X$ such that $\map f x = \map g x = +\infty$.

We avoid this technicality by introducing the sets $A$ and $B$, as subtraction is well-understood on $A \cap B$.

We write $f \cdot \chi_{A \cap B} - g \cdot \chi_{A \cap B}$ instead of $\paren {f - g} \cdot \chi_{A \cap B}$ for this reason.

Also see

 * Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal