While you are at it, you could allow the value $+\infty$ as well...
The resulting object is a random variable from a probability space $(\Omega,\mathcal{F},P)$ to a bona fide measurable space $(E,\mathcal{E})$, in this case $E=\mathbb{R}\cup\{-\infty,+\infty\}$ and $\mathcal{E}$ the Borel $\sigma$-algebra of $E$ endowed with its usual topology (roughly speaking, this is equivalent to seeing $E$ as the closed interval $[0,1]$ endowed with its subspace topology). A description of $\mathcal{E}$ is that $A\subset E$ is in $\mathcal{E}$ iff there exists $B\in\mathcal{B}(\mathbb{R})$ and $I\subset\{-\infty,+\infty\}$ such that $A=B\cup I$.
So you are squarely in probability theory in the most standard sense of the term.
Rereading your post, I feel I should mention that $E(X|X\in\mathbb{R})$ need not be well defined, for the same reason that a real valued random variable need not be integrable.
Let us consider more closely the question about space-filling curves. The Peano curve and the Hilbert curve, and several other variations of them, have parametrizations $[0,1]\to[0,1]^2$ that actually take the 1-dimensional Lebesgue measure on $[0,1]\, ,$ $\mathcal{L} _1\, ,$ to the 2-dimensional Lebesgue measure on $[0,1]^2\, ,$ $\mathcal{L} _2\, ,$ by push-forward. Notably, the Peano curve $\gamma: [0,1]\to[0,1]^2$ is originally defined in terms of ternary expansions of real numbers, which makes it particularly simple to check the independence of the coordinates. The Hilbert curve also admits a (slightly less) simple description in terms of binary expansions. Here are the details; I'll try to give you an abstract-nonsense-friendly description.
Let's start by some well known facts. The map that takes a sequence $a:=(a_1,a_2,\dots)\in\ 3^\mathbb{N}$ (here $3:=\{0,1,2\}$ and $\mathbb{N}$ is the set of positive integers) into its value as ternary expansion, namely $\mathrm{v}(a):= \sum_ {n\ge1} 3^{-n}a_ n\in[0,1]$, is a continuous surjective map (and bijective, up to removing a certain countable, null subset $D$ of the domain) :
$$\operatorname{v}:3^\mathbb{N}\to[0,1]\, .$$
Also, this map takes the measure $\mathbf{m}$, product of countably many copies of the uniform probability measure on $3$ to the measure $\mathcal{L} _1$ ("the base 3 digits of a real number are independent and uniformly distributed". Any other base of course works as well).
Now, at the level of the ternary sequences we have the nice and simple Cantor bijection $3^\mathbb{N}\to 3^\mathbb{N}\times 3^\mathbb{N}$
"split the sequence of digits into the sequence of odd-position digits and the sequence of even-position digits"
$$C:3^\mathbb{N}\ni (a_n)\mapsto \big(\, (a_{2n-1}), (a_{2n})\, \big) \in 3^\mathbb{N}\times 3^\mathbb{N} $$
which is easily seen to be a compact metric space homeomorphism that takes the measure $\mathbf{m}$ into the product measure $\mathbf{m}\otimes\mathbf{m}$. Note that this homeomorphism does not pass through the quotient map $\mathrm{v}$, for in general sequences with the same value do not produce sequences with the same value by extraction of a subsequence. However, removing the above mentioned countable set $D$ the map $\mathbf{v}$ becomes bijective and you do have correspondingly a bi-measurable, a.e. defined (or everywhere but non-continuous) map of the unit interval to the unit square that takes the measure $\mathcal{L} _1$ to the measure $\mathcal{L} _2$. All that is quite standard. Now your question cames quite naturally, as a request for a commutative square:
Can we find another measure preserving homeomorphism
$\Gamma:3^\mathbb{N}\to
> 3^\mathbb{N}\times 3^\mathbb{N}$ that
induces a map $\gamma:
> [0,1]\to[0,1]\times[0,1]$ through the map $\mathbf{v}$, that is, such that $\gamma\circ\mathbf{v}=(\mathbf{v}\times\mathbf{v})\circ\Gamma$ ?
Since $\mathbf{v}$ is a quotient map, this map $\gamma$ will be automatically a continuous surjection, that also takes $\mathcal{L} _1$ to $\mathcal{L} _2$. The answer is yes, and this is Peano's construction (he was not interested in the measure-theoretic property, but this also follows immediately from the definition). It's the way he constructed his example in the celebrated paper dated 1890 on Mathematische Annalen, " Sur une courbe, qui remplit toute une aire plane ".
Here's Peano definition of the map $\Gamma$: extract as before the sequence of the odd-position digits and the even-position digits, but first invert every odd-position digit whenever there are an odd number of odd even-position digits before it, and invert every even-position digit
whenever there are an odd number of odd odd-position digits before it. Here "invert" just means taking $x\in3$ to $2-x$, that is $0$ to $2$, $1$ to $1$, $2$ to $0$. Translating this definition in a formula, is not difficult to check it defines a homeomorphism of the form $\Gamma=C \circ \phi$, compatible with the map $\mathbf{v}$. The measure property is quite obvious, since the "inverting digit map" $ \phi:3^\mathbb{N}\to 3^\mathbb{N}$ is clearly an involutory preserving measure homeomorphism.
For the Hilbert curve, the digit description has to be done in terms of binary representation, and it is slightly less simple (I have it written somewhere and will look for it and quote here at request) but everything works as well.
Let me finish with an historical note. Of course, what is not easy in the short Peano's paper is to understand what's going on geometrically. He made no picture in this paper, although the graphical iterative construction was perfectly clear to him, and was with all probability his starting point —he made an ornamental tiling showing a picture of the curve in his home in Turin. His choice to avoid any appeal to graphical visualization was no doubt motivated by a desire for a well-founded, completely rigorous proof owing nothing to pictures, in the spirit of the program of arithmetization of analysis. In the conclusion of his paper he observed incidentally that the same construction may be made with all odd basis, and even basis too, although in the latter case, by slightly more complicated formulae (hence less elegant from his viewpoint). In order to make Peano's example more accessible to the mathematical community, a couple of years later Hilbert wrote on the same journal the very clear geometric construction that we know. He chose the Peano construction in base 2 because it is simpler from the graphical point of view.
Best Answer
$Y(X)$ doesn't mean anything. You can't define the composition of random variables. What you can do is compose a random variable $X$ by a measurable function $f$ (provided the $\sigma$ algebras are the same) : $f(X)$.
So in your example, there are two different objects, measurable functions and random variables :
-the measurable functions $f$ from $(A,\sigma_A)$ to $(B,\sigma_B)$ and $g$ from $(B,\sigma_B)$ to $(C,\sigma_C)$. Since B uses the same $\sigma$-algebra, the function $g \circ f$ is measurable from $(A,\sigma_A)$ to $(C,\sigma_C)$.
-the random variable when you add a probability distribution to the measurable spaces. So if you add $P_A$ to $(A,\sigma_A)$, the measurable function $f$ from $(A,\sigma_A)$ to $(B,\sigma_B)$ induces a random variable we can write $X$. Now since we also have a measurable function $g \circ f$ from $(A,\sigma_A)$ to $(C,\sigma_C)$, it also induces another random variable that we can write $X'$ or more usually $g(X)$. And if you add $P_B$ to $(B,\sigma_B)$, function $g$ induces a random variable we'll write $Y$.
But the composition $Y(X)$ doesn't makes any sense.