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If $\mathscr{G}$ is a sub-$\sigma$-algebra of $\mathscr{F}$ and if X $\in L^1 (\Omega, \mathscr{F}, P)$ and if Y $\in L^1 (\Omega, \mathscr{G}, P)$ and $E(X;G) = E(Y;G)^{*} \forall G \in \mathscr{I}$ where $\mathscr{I}$ is a $\pi$-system containing $\Omega$ s.t. $\sigma(\mathscr{I}) = \mathscr{G}$, then $E(X;G) = E(Y;G) \forall G \in \mathscr{G}$.

My attempt:

$E(X;G)$ and $E(Y;G)$ are finite measures that agree on a pi-system. They are finite because you can plug in $\Omega$ in each and say the output is finite X $\in L^1 (\Omega, \mathscr{F}, P)$ and Y $\in L^1 (\Omega, \mathscr{G}, P)$ and $\Omega \subseteq \mathscr{I} \subseteq \sigma(\mathscr{I}) = \mathscr{G} \subseteq \mathscr{F}$. Thus by uniqueness lemma, they agree on the sigma-algebra generated by the Pi-System. QED…? Is this right?

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Oh also, they are measures because they are $\sigma$-additive and plugging in $\emptyset$ makes them zero.

*I’m not sure but I’m guessing this means $\int_G X dP = \int_G Y dP$

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If $X\geqslant 0$, then we may argue that

$$

\mathscr{G}\ni G\mapsto \mathrm{E}[X;G] \quad\text{and}\quad \mathscr{G}\ni G\mapsto \mathrm{E}[Y;G]

$$

are measures that agree on the $\pi$-system $\mathscr{I}$ and thus also on the sigma-algebra $\sigma(\mathscr{I})=\mathscr{G}$. Exactly, as you have written.

Now for a general $X\in L^1(\Omega,\mathscr{F},P)$ we have

$$

\mathscr{G}\ni G\mapsto \mathrm{E}[X^+;G]+\mathrm{E}[Y^-;G]

$$

and

$$

\mathscr{G}\ni G\mapsto \mathrm{E}[Y^+;G]+\mathrm{E}[X^-;G]

$$

are measures that agree on the $\pi$-system $\mathscr{I}$ and hence…

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