conditional probabilty and the sample space BACKGROUND
Allow that an experiment 1 and 2 are defined by a probability space triple $(\Omega_1, \mathcal{F}_1, P_1)$, and $(\Omega_2, \mathcal{F}_2, P_2)$, respectively [1]. 
Allow that an experiment 2 is in all ways identical to experiment 1, except that there is one additional condition imposed. This condition is such that certain elements (also known as outcomes) of sample space $\Omega_1$ are not present in sample space $\Omega_2$.
QUESTION
In your opinion, if I am writing a journal article, should I define each probability space separately---like i have done above---; or can I simply defined the experiment number 2 by defining a predicate on probability space 1?
MY OPINION
My opinion is as follows. I know that "...the sample space of an experiment or random trial is the set of all possible outcomes or results of that experiment. [1]" I know that certain outcomes in $\Omega_1$ are not in $\Omega_2$. Therefore, it is strictly wrong to state that the sample space of experiment 2 is $\Omega_1$. But perhaps I am being overly pedantic
BIBLIOGRAPHY
[1]
https://en.wikipedia.org/wiki/Probability_space
 A: The events of the second experiment will be conditioned on some observed evidence. This observation results in some outcomes discarded and now we are in the restricted sample space whose outcomes are renormalized to add up to 1.
Let us take an example here.
Two cards are drawn (without replacement) from a deck of 52 cards. Let A = event that first card is a heart. Let B be the event that the second card is red. The sample space is 52 cards. Its subsets are various events like A, B, etc. Event A is the set {13 cards of heart}. Event B is the set {26 red cards}. We can calculate their probabilities as P(A) = 1/4 and P(B) = 1/2.
What about P(A|B)? P(A|B) is the probability of the event A given that the event B has already happened. If event B has already occurred, the restricted sample space has 51 cards, 25 of which are red.
The definition of P(A|B) = P(A and B)/P(B).
Here, P(B) normalizes/transforms the probabilities from the original sample space to the restricted sample space.
P(A and B) = (13/52).(25/51)
Hence, P(A|B) = (13/52).(25/51) / (1/2) = 25/104 which is slightly lower than 1/4 because one red card less in the restricted sample space reduces the chances of getting a heart.
So, are we better off using a separate sample space of 51 cards, 25 of which are red, for the second experiment? No, because we do not know if the removed red card is a heart or a diamond.
What if B too is an event that a heart was removed? We then have a restricted sample space of 51 cards, 25 of which are red, and 12 of which are hearts. We can now use this as our sample space. But, remember, we still have to deal with the  the problem of black cards are more likely than red cards. When we are dealing with marginalization using conditioning, "the choice of how to divide up the sample space is crucial: a well-chosen partition will reduce a complicated problem into simpler pieces, whereas a poorly chosen partition will only exacerbate our problems, requiring us to calculate n difficult probabilities instead of just one!" - Blitzstein, Hwang.
So, if the experiment is the same with one sample space being a restricted form of another, we are better off using the same sample space with the restriction applied rather than defining a new one.
References:
[1] Blitzstein J.K., Hwang J. - Introduction to Probability-CRC (2015)
