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If $X$ is some $\mathcal F$-measurable random variable, and we have two sigma algebras $\mathcal A$ and $\mathcal B$ that satisfy $\mathcal A \subset \mathcal F$ and $\mathcal B \subset \mathcal F$, and that $\int_A X dP = 0$ for all $A\in \mathcal A$ and for all $B\in \mathcal B$, it it true that $\int_D X d P = 0$ for all $D\in \sigma[\mathcal A \cup \mathcal B]$?

This kind of reminded me of Caratheodory's Extension Theorem extending a pre-measure on some collection of sets to an actual measure on the $\sigma$-algebra generated by that collection. I was thinking we could make some measure $\mu(D)$ defined by these integrals, that would extend to be $0$ on the $\sigma$-field. However, I don't think $\mu=0$ would have any impact on the original integral being $0$, right?

This is something that came up when I was trying to prove that $\int_D (X-Y) d P = 0$ for all $D \in \mathcal F(X-Y) := (X-Y)^{-1}(\mathcal B)$ if we are given that $\int_D (X-Y) d P = 0$ for all $D \in \mathcal F(X) \cup \mathcal F(Y)$. Intuitively this seems true, and I've tried to come up with counterexamples with no avail. So, I conjectured the above question. More generally, is it true if the integrals are 0 on all sets in $\cal A_1, \cal A_2, \ldots, \cal A_n$, it's also zero on the $\sigma$-algebra containing these sub-$\sigma$-fields?

D.R.
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Not true. If $\mathcal A=\{\emptyset, \Omega, A,A^{c}\}$ and $\mathcal B=\{\emptyset, \Omega, B,B^{c}\}$ then the hypothesis is satisfied when $\int _A XdP=\int _B XdP=\int _{\Omega} XdP=0$. These conditions do not imply that $\int _{A\cup B} XdP=0$.

Let $X$ take the values $-3,-2,-1,1,2,3$ with probabilities $\frac 2 {34},\frac 9 {34},\frac 6 {34},\frac 6 {34},\frac 9 {34},\frac 2 {34}$ respectively. Let $A=\{-3,1\}$ and $B=\{-3,3\}$. You can verify that this is provides a counter-example.

Kavi Rama Murthy
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  • sorry, I can't think of one. I'm doodling step functions and making them go to 0, but can't find an explicit example. Can you help me? – D.R. Feb 05 '20 at 06:07
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    @D.R. Look at my answer now. – Kavi Rama Murthy Feb 05 '20 at 06:30
  • Ok thank you. Unfortunately, my proof direction for the claim in the third paragraph is now wrong. Do you have any further ideas as to how to prove that claim? – D.R. Feb 05 '20 at 06:36
  • You say 'intuitively this seems true' but your intuition has failed here. @D.R. – Kavi Rama Murthy Feb 05 '20 at 06:44
  • well for the claim in the 3rd paragraph I'm more sure it's true because I was tasked with proving it in a class. https://math.stackexchange.com/questions/666843/if-exy-y-almost-surely-and-eyx-x-almost-surely-then-x-y-almost-surel has a proof using conditional expectations, and I wanted to try to prove it solely using these methods. – D.R. Feb 05 '20 at 07:09