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

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

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

Suppose that $\mu \restriction_{\mathcal G}$, the restriction of $\mu$ to $\mathcal G$, is $\sigma$-finite.


Let $f, g: X \to \overline{\R}$ be $\mathcal G$-measurable functions.

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

$\displaystyle \int_G f \, \mathrm d \mu = \int_G g \, \mathrm d \mu$


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


Proof


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