# Equality of integrals on intersection stable generator implies equality on entire $\sigma$-algebra

Let $X$ be a random variable on a probability space $(\Omega, \mathcal A, P)$ such that $E|X|<\infty$. Suppose $\mathcal G \subset \mathcal A$ is a sub-$\sigma$-algebra with intersection-stable generator $\mathcal M$ such that $\Omega \in \mathcal M$. Let $Y\in L^1(\Omega, \mathcal G, P)$. Then $$\int_AX dP =\int_AY dP$$ for every $A\in \mathcal M$ implies that the above equality holds for every $A \in \mathcal G$.

How do I approach this problem? The first thing that jumps to mind whenever we have an intersection-stable generator is that its generated Dynkin-system coincides with its generated $\sigma$-algebra. Does this imply that we can write $A\in \mathcal G$ as the disjoint union of sets that lie in $\mathcal M$ or whose complement lies in $\mathcal M$?

I'm thankful for any help.

• – BCLC Sep 18 '15 at 13:23

If $\mathcal D:=\{A\in\mathcal G\mid \int_AXdP=\int_AYdP\}$ then it can be proved that $\mathcal D$ is a Dynkin-system.
This with $\mathcal M\subseteq\mathcal D$ so that $\mathcal D$ will contain the Dynkin-system generated by $\mathcal M$.
Since $\mathcal M$ is intersection-stable this collection will coincide with the $\sigma$-algebra generated by $\mathcal M$ wich is $\mathcal G$ (as you said).
So $\mathcal G\subseteq\mathcal D$ as was to be shown.
• does this property hold for non sigma-finite measures? (i.e., if we replace $P$ by an arbitrary measure $\mu$) – user2139 May 3 '18 at 6:43
• @user2139 Yes. For integrable functions $X,Y$. – drhab May 3 '18 at 7:14