Why not defining random variables as equivalence classes?

The usual definition of a random variable (or random element) is that of a measurable function $X : (\Omega, \mathcal{F}, P) \rightarrow (\Omega', \mathcal{F}')$. Now I am not aware of any property/theorem that depends on the specific values of $X$ for every $\omega \in \Omega$. In particular any other $P$-almost surely equal random variable $X'$ is generally considered as equivalent to $X$ for all practical purposes.

So is there a good reason not to define random variables as equivalent classes rather than laboriously precising each time that such or such statement is true almost surely, that such or such sequence converges almost surely, that such or such object is unique almost surely, etc ? As a comparison defining $L^p$ spaces as spaces of equivalent classes of almost everywhere equal functions helps a lot in simplifying the phrasing of the theory.

So are there some interesting/complex cases where we would really need to keep the distinction between almost surely equal random variables?

Edit

In agreement with @Pedro Tamaroff's comment I'm removing the last addendum to this question and opening a new one.

• First examples coming to mind concern the almost sure properties of realizations of random processes indexed by uncountable sets, say the almost sure Hölder continuity of the paths of Brownian motion $(B_t)$. If one allows to modify each random variable $B_t$ on a null set, the resulting path $t\mapsto B_t(\omega)$ may become ugly on an event of positive probability.
– Did
Nov 21, 2016 at 23:49
• Indeed, if $B_t$ were defined as a one-parameter family of equivalence classes, then a question like "is $t \mapsto B_t(\omega)$ continuous for almost every $\omega$" becomes ill defined, since the answer may depend on the choice of representatives. This is the idea behind a modification of a stochastic process. Nov 22, 2016 at 0:01
• @Did: I think this should be an answer. My knowledge of post-undergrad probability is nil, but that is a very convincing example (unlike the only answer so far, which I don't even understand). Nov 22, 2016 at 0:05
• OP: So, in the end, you modified substantially the question after you received answers and after you have been signalled such modifications are to be avoided because they have the undesirable effect of making previous answers off-topic. A reversal to the previous version of the question seems desirable.
– Did
Dec 5, 2016 at 12:44
• @Burakumin Please, do not modify a post after answers have been provided with edits that render such answers incomplete, incoherent or altogether incorrect. Ask a new question instead. Regards,
– Pedro
Dec 5, 2016 at 17:11

One of the first classes of examples coming to mind where this matters concerns the almost sure properties of realizations of random processes indexed by uncountable sets, say the almost sure Hölder continuity of the paths of Brownian motion $(B_t)$. If one allows to modify each random variable $B_t$ on a null set, the resulting paths $t\mapsto B_t(\omega)$ may become ugly for every $\omega$ in an event of positive probability.

Edit: Regarding "ugly" above, user @tomasz mentioned a useful point in a comment below, which I now reproduce: if one allows to modify each random variable on a null set, the supremum of an arbitrary (uncountable) family of measurable functions need not be measurable, not even if the functions are almost everywhere zero (say, indicators of points).

• What "ugly" means ? Wouldn't we have something like: « For almost every $\omega$, $t \mapsto B_t(\omega)$ is almost everywhere continuous. » ? If this is the case that does not sound as a big loss. Nov 22, 2016 at 9:09
• This is indeed "a big loss" if one insists that the paths are continuous, almost surely. And one often does. (Did you vote to delete this, by any chance?)
– Did
Nov 22, 2016 at 10:28
• I think a big issue here is that the supremum of an arbitrary (uncountable) family of measurable functions need not be measurable, not even if the functions are almost everywhere zero (say, indicators of points). Nov 23, 2016 at 11:13
• @tomasz Yes. Point added to my answer (if you disagree, just yell). :-)
– Did
Nov 23, 2016 at 11:23
• @Burakumin Why are you staying silent? You were asked at least one precise question, if only for good manners you could see fit to address it...
– Did
Nov 23, 2016 at 11:24

Although every finite moment of a variate $X$ will be equal to the corresponding moment of the two "equivalent" variate $X^\star$, one might consider the range of a distribution as an interesting property. If you consider the range to be interesting, then consider the following two distributions, both derived from an underlying uniform random $U$ on $[0,1]$: $$X: \begin{array}{lc} X = & \left\{ \begin{array}{cl} U & U \mbox{ irrational}\\ -U & U \mbox{ rational} \end{array}\right.\\ X^\star = &U \end{array}$$ The variate $X$ is almost surely equal to $X^\star$ but the range of $X$ is $[-1,1)$ whilst the range of $X^\star$ is $(0,1)$.

• I don't know if that's a compelling example; to many it would just suggest that range is not in fact interesting, and one should instead look at essential range, which is well-defined with respect to a.s. equality, or even equality of distribution. In this case both random variables have $[0,1]$ as their essential range. Nov 22, 2016 at 0:04