Given two random variables X,Y with $E |X|<\infty$, $E |Y|<\infty$ and the equality
$$\int_{D}X dP=\int_{D}Y dP, D\in\mathcal{D}$$
where $\mathcal{D}$ is some sigma algebra on the background space.
What are the sufficient conditions so that we can conclude that $X$ and $Y$ are almost surely equal?
Likewise, if we have two random variables that are almost surely equal, do we have the equality in the integrals above in general?
Best Answer
This is quite a standard argument in measure theory. You don't need any additional assumptions (if I understand it correctly that you suppose the equality holds for all $D$ in the $\sigma$-algebra).
You can easily reduce the problem to the setting $$\int_D X dP=0 $$ for all $D\in \mathcal{D}$ by taking the difference of $X$ and $Y$ (both sides are finite by your assumption) and we will show that $X=0$ almost surely in that case. Let $A_n:=\{\omega:X(\omega)>\frac{1}{n}\}$. If for any $n\in \mathbb{N}$ the set $A_n$ has nonzero probability we obtain the contradiction $$\int_{A_n} X dP\geq \frac{P(A_n)}{n}>0.$$ Therefore, $P(A_n)=0$ for all $n$. Furthermore, we have $$A:=\{\omega:X(\omega)>0\}=\bigcup_{n\in \mathbb{N}}A_n$$ and by the $\sigma$-subadditivity this implies $P(A)=0$. With the same argument we also obtain $P(\{\omega:X(\omega)<0\})=0$ and, thus, $P(\{\omega:X(\omega)\neq 0\})=0$ which is equivalent to $X= 0$ almost surely.
For the other direction note that you can always disregard sets of measure zero in interals due to the convention $\infty \cdot 0=0$, i.e., you can always take the integrals only over the set where the two functions coincide and there the integrals are trivially equal.