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

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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 $G \in \GG$:

$\displaystyle \int_G f \rd \mu = \int_G g \rd \mu$


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


Proof


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