As stated in the title,
How to prove that $$L^0( \mathbb P | \mathcal F_0 ; || \cdot ||_{\mathbb R^d} ) \subset L^0( \mathbb P | \mathcal F_1 ; || \cdot ||_{\mathbb R^d} ) $$
where the space $ L^0( \mathbb P | \mathcal F_0 ; || \cdot ||_{\mathbb R^d} ) $ is given by the equivalence class of all measurable random variables, given the probability measure $\mathbb P$ restricted to the filtration $\mathcal F_t $. Apparently, it should be enough to prove:
$(\Omega, \mathcal F)$, $(S, \mathcal S)$, two measurable space, $\mathcal F_t$ a filtration on $\mathcal F$,
$X \colon \Omega \to S$, an $\mathcal F, \mathcal S $ measurable function, $Y \colon \Omega \to S$, an $\mathcal F_t, \mathcal S$ measurable function, and such that $\exists A \in \mathcal F \colon \mathbb P(A) = 1$ as well as $A \subset \{ X \equiv Y \}$. Then, it implies $ X$ is a $\mathcal F_t, \mathcal S $ measurable function.
So my question is how are these two theoremes related and how to prove the former?
My attempt.
In the proof of the later, I am stuck at the step where I have to use the probability measure. Indeed, I see that the preimage of $X$ should lie on the filtration, the problem is that I am not able to express the idea that if the preimage is on $\mathcal F \setminus F_t$ then the image has measure $0$.
after discussing it with other people, the whole point comes down to using the stochastic bases.
So, first, one has that the set $A$ I have given the definition above is in the filtration $\mathcal F_t$. This is because of the stochastic bases where all sets in the kernel of the measure are in the filtration. Then, one can partition $\{ X \in B \}$ upon $A$, giving the union of two sets: one is in the filtration, and the other one is of measure zero, thus in the filtration (same argument).
Using the same method, we can show the inclusion of spaces.