Measurable 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 $\Sigma$.

Suppose that $\mu \restriction_\GG$, the restriction of $\mu$ to $\GG$, is $\sigma$-finite.

Let $f, g: X \to \overline \R$ be $\GG$-measurable 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
First, assume that $f$ and $g$ are $\mu$-integrable.

Observe:

For each $n \in \N_{>0}$:

Swapping $f$ and $g$, we also have:

Thus:

Therefore:

Hence the result, by definition of almost-everywhere equality.

Now, consider general $f$ and $g$.

Recall that $\mu \restriction_\GG$　is $\sigma$-finite.

That is, there is an exhausting sequence $\sequence {E_n}_{n\in\N} \subseteq \GG$ such that:
 * $\map \mu {E_n} < \infty$

We define $\sequence {A_n}_{n\in\N} \subseteq \GG$ by:
 * $A_n := E_n \cap \set { \size f \le n} \cap \set {\size g \le n}$

Then $\sequence {A_n}_{n\in\N}$ is also an exhausting sequence.

Let $f_n := f \chi_{A_n}$ and $g_n := g \chi_{A_n}$.

Then $f_n$ and $g_n$ are $\mu$-integrable so that:
 * $\forall n \in \N : \map \mu {\set {f_n \ne g_n} } = 0$

On the other hand:
 * $\set {f_n \ne g_n} = \set {f \ne g} \cap A_n$

Therefore:

Also see

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