Question on $\mathcal{L}(\Omega, \mathcal{A}, P) \subseteq \mathcal{L}(\Omega, \mathcal{F}, P)$ Let $\mathcal{A}\subseteq \mathcal{F}$ be sigma-algebras, then $\mathcal{L}(\Omega, \mathcal{A}, P) \subseteq \mathcal{L}(\Omega, \mathcal{F}, P)$. But then for any $f\in \mathcal{L}(\Omega, \mathcal{A}, P)$, this obviously means that $f\in \mathcal{L}(\Omega, \mathcal{F}, P)$. Let's say there exists an event $B \in \mathcal{F}$ but $B \notin \mathcal{A}$
Surely I can then not evaluate $\int_{B}f(x)dx$ since $f\in \mathcal{L}(\Omega, \mathcal{A}, P)$ and $B \notin \mathcal{A}$, in other words $f ^{-1}(B)=\varnothing$
but $f$ is also in $\mathcal{L}(\Omega, \mathcal{F}, P)$ so surely this is a contradiction?
 A: You definitely can evaluate $\int_Bfdx$, (I assume by $\mathcal L(\Omega,\mathcal F,P)$ you are referring to integrable functions on the measure space). I am not sure why you think you can't. It does not make sense to talk about the integral of $f$ over $B$ if all you knew of was the smaller $\sigma$-algebra, but you know that it exists in a larger measure space with the same (extended) measure, so the integral is perfectly well defined.
By saying that $f\in\mathcal L(\Omega,\mathcal A,P)$ all you are saying is that $f:\Omega\to \mathbb R$ (or onto some other topological space) is a function such that $f^{-1}(A)\in\mathcal A$ for any Borel set $A$ in $\mathbb R$, and $\int_\Omega|f|dx<\infty$. You are not saying that $f$ can only be evaluated on sets in $\mathcal A$. As $B\in \mathcal F$, we know that $P(B)$ is a well defined entity, and $\int_Bfdx$ is just the standard Lebesgue integral.
I have no idea what you mean by saying that $f^{-1}(B)=\emptyset$. $B$ is a subset of $\Omega$, and $f$ maps from $\Omega$ to $\mathbb R$. You can only talk about the forward image of $B$ under $f$, and there is absolutely no reason why it must be empty.
A: I think that you can evaluate $\int_{B}f(x)dx$, because $f\in \mathcal{L}(\Omega, \mathcal{F}, P)$.
I would say that you can not conclude from $B \notin \mathcal{A}$ that $\int_{B}f(x)dx$ can not be evaluated.
Still, the other way around, it holds, I would say, i.e. if $\int_{B}f(x)dx$ can not be evaluated, then $B \notin \mathcal{A}$.
