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
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.

Also see

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