What is the probability space of typical real univariate probability distributions? Postscript to the question below. In trying to learn from the answers below, all of which I am grateful for, I read a historical article on the origins and legacy of Kolmogorov's Grundbegriffe.  This article helped me understand what basic things people were struggling with when this theory was developed. In particular, the long-term trend towards abstraction and foundation in terms of measure theory, and the early-days focus on the connection between the real world and the probabilistic model.  I then re-read the answers and comments. I made a comment that started

We can choose $Ω=\Re$ because the domain of the distribution function is $\Re$.

This is wrong because the domain of the distribution function is not necessarily mentioned in the declaration of the probability space.  I made the convention that random variables $X: \Omega \rightarrow \Re$.  So the domain of the distribution function is $\Re$ by my convention, but that doesn't have anything to do with the probability space. $\Omega$ is a kind of index set.  Suppose we are reasoning about the saturation of the color red in grapes.  In that case we are thinking about say a color level in $S=[0,255)$. Nowhere in the definition of a probability space $(\Omega,\mathcal A,P)$ to support reasoning about $S$ do we need to specify $S$.  We do need to demonstrate that there is a 1-1 mapping between $\Omega$ and $S$, i.e. that $\Omega$ can enumerate $S$. Once we have "built" $(\Omega,\mathcal A,P)$, we can put it to work and re-use it for any $S$ which $\Omega$ can enumerate. The probability space $(\Omega,\mathcal A,P)$ is a kind of indexing structure. That for me is the key realization.  The key cognitive error comes from labelling $\Omega$ as the sample space, and $\mathcal A$ as the event space.  The common sense meaning of those terms implies a connection with the actual samples being reasoned about, when that does not have to be the case.  A far less misleading terminology would be to label $\Omega$ as the sample index space or just index space, and $\mathcal A$ as the index set space.  This kind of thing is clearly understood in programming languages, where if I have an array $A$, then $(i,j)$ is an index and I don't confuse $(i,j)$ with $A[i,j]$, and I don't confuse the purpose of arrays with the purpose of array indices, but in some contexts I can identify $A[i,j]$ with $(i,j)$.
Short version of the question:  How do we formally and correctly define the probability space of the reals which supports the definition of the typical/usual univariate continuous probability distributions, such as uniform and exponential?
Short restatement of the core question that I have:  I am hung up on p. 3 section 1.1B of the KPS text.  They start with an unspecified probability space $(\Omega,\mathcal A,P)$.  Two distinct random variables $V$, $V \in Exp(\lambda)$ and $V \in U[a,b]$, are said to have distribution functions $F_V=P_V((-\infty,x))=P(\{\omega \in \Omega: V(\omega)<x\})$.  These are distinct and solved separately as $F_{U[a,b]}(x) = \mathcal H(x-a) \mathcal H(b-x) \frac{x-a}{x-b} + \mathcal H(x-b)$ and $F_{Exp(\lambda)}=\mathcal H(x) (1-e^{-\lambda x})$, where $\mathcal H(x \geq 0) = 1, \mathcal H(x<0)=0$. My key question is:

*

*What is a solution for the $P$ shared by $X$ and $Y$?

Note: Here are some similar questions on Math Stack Exchange

*

*Probability space of a Gaussian, unanswered, from 2016.

*What are the sample spaces when talking about continuous random variables?, asked 9 years ago and answered as $[0,1]$.  The accepted answer starts out by saying "You can take it to be a subset of $\Re$ or, more generally, $\Re^n$."  But then the solver gets to $[0,1]$.

Comment: I was mistakenly assuming that the text above was taking $\Omega=\Re$ because I saw a similar statement somewhere to the effect of saying "for purposes of discussion let's say the sample space for continuous random variables is $\Re^d$".  The cited answer to 2nd question above starts that way but then gets to $[0,1]$.   So: I now understand that the $[0,1]$ is the "best fit" sample space, along with Lebesgue measure.  So the "right" probability space that I was looking for is the Steinhaus space $([0,1],\mathscr B([0,1]), \mu)$ where $\mu$ is the Lebesgue measure restricted to $[0,1]$.  99.999% of my confusion came from

*

*Not recognizing that $[0,1]$ is a "big enough" space to enumerate the domain of a continuous map into $\Re$.  So it's "as good as" $\Re$.

*Making the assumption that the convention, was, somehow somewhere, to identify the sample space for $d$-dimensional continuous random variables with $\Re^d$, when the "best fit" answer is $[0,1]^d$.

Longer version of the question:
Following this text,
Let $\Omega$ be a nonempty set, the sample space.
Let set $\mathcal F$ of subsets of $\Omega$ be a $\sigma$-algebra so that

*

*$\Omega \in \mathcal F$

*$\Omega \setminus F \in \mathcal F$ if $F \in \mathcal F$

*$\bigcup_{n=1}^{\infty} F_n \in \mathcal F$ if all $F_i \in \mathcal F$
Let $P: \mathcal F \rightarrow [0,1]$ be a probability measure so that

*

*$P(\Omega) = 1$

*$P(\Omega \setminus F) = 1-P(F)$

*$P(\bigcup_{n=1}^{\infty} F_n) = \sum_{n=1}^\infty P(F_n)$
We call the triple $(\Omega, \mathcal F, P)$ a probability space.
Suppose $X:\Omega\rightarrow \Re$.  We say $X$ is a random variables if $\{\omega \in \Omega : X(\omega) \leq a\}$ is in $\mathcal F$ for every $a \in \Re$.
Then the probability distribution function $F_X : \Re \rightarrow \Re$ is defined for all $x \in \Re$ as
$$F_X(x) = P(\{\omega \in \Omega : X(\omega) < x\})$$
Note that $P$ appears unsubscripted in the definition of $F_X$.  $P$ does not depend on the particular random variable $X$ whose distribution we are defining.  So in that sense it should be possible for the same probability space $(\Omega, \mathcal F, P)$ to underly probability distribution function constructions for multiple distinct random variables $X$ and $Y$, $X \neq Y$, for the same probability space.
For example, let
$$\Omega = \{0,1\}$$
$$\mathcal F = \{\emptyset, \{0\}, \{1\}, \{0,1\}\}$$
$$P = \begin{cases}
\emptyset &\mapsto& 0 \\
\{0\} &\mapsto& \frac{1}{2} \\
\{1\} &\mapsto& \frac{1}{2} \\
\{0,1\} &\mapsto& 1
\end{cases}$$
Let $X,Y: \Omega\rightarrow \Re$ and be random variables fully defined by
$$X = \begin{cases}
0 &\mapsto& 17 \\
1 &\mapsto& 17
\end{cases}$$
$$Y = \begin{cases}
0 &\mapsto& 42 \\
1 &\mapsto& 42
\end{cases}$$
Then the probability distributions of $X$ and $Y$ are
$$F_X(x) = P(\{\omega\in\Omega:X(\omega)<x\}) = \begin{cases}
 x < 17 &\mapsto& 0 \\
x \geq 17 &\mapsto& 1
\end{cases}$$
$$F_Y(x) = P(\{\omega\in\Omega:Y(\omega)<x\}) = \begin{cases}
 x < 42 &\mapsto& 0 \\
x \geq 42 &\mapsto& 1
\end{cases}$$
Clearly $X \neq Y$ and $F_X \neq F_Y$.  In the above discrete example, if I understand the language correctly, there is a single probability space $(\Omega,\mathcal F,P)$ with a single probability measure $P$ which underlies or supports two distinct probability distributions $F_X$ and $F_Y$ for two distinct random variables $X$ and $Y$.
Now let $(\Omega, \mathcal F, P)$ be a probability space underlying random variables $X$ and $Y$ where:

*

*Random variable $X: \Omega \rightarrow \Re$ is such that $X$ has the uniform distribution $F_X: \Re \rightarrow [0,1]$ such that

$$F_X(x) = P(\{\omega\in\Omega:X(\omega)<x\}) = \begin{cases}0  &:& x < a \\
\frac{x-a}{b-a}  &:& a \leq x \leq b \\
1 &:& b < x
\end{cases}$$

*

*Random variable $Y: \Omega \rightarrow \Re$ is such that $Y$ has the exponential distribution $F_Y: \Re \rightarrow [0,1]$ such that

$$F_Y(x) = P(\{\omega\in\Omega:Y(\omega)<x\}) =  \begin{cases}0  &:& x < 0 \\
1-e^{-\lambda x} &:& x \geq 0
\end{cases}$$
Also, per comment below, one distribution can be supported by multiple probability spaces.  (The key understanding here for me is that probability space and probability distribution are separate constructions.)
My questions are (and some answers that I take from my reading of the solutions below):
Q1. Is $(\Omega, \mathcal F, P) = (\Re, \mathcal B(\Re), \mu)$ where $\mathcal B(\Re)$ is the Borel set of the reals and $\mu$ is the Lebesgue measure a probability space which underlies $X$ and $Y$?  Answer: No, but the Steinhaus $([0,1], \mathcal B([0,1]), \mu)$ is good.
Q2. Is it correct to call $(\Re, \mathcal B(\Re), \mu)$ the standard probability space of the reals?  Is there some other standard notation or language for the probability space underlying the usual continuous probability distributions?  Answer: No, but the Steinhaus space is a standard space in the Wikipedia sense.
Q3. Is it correct to say that the notion of probability space is independent of and complementary to the notion of probability distribution, and that the notion of probability distribution is always associated with a particular random variable $X$ presented with a supporting probability space $(\Omega, \mathcal F, P)$?  Answer: Kind of.  One distribution can be accompanied by many probability spaces.  One probability space can be accompanied by many distributions.  I'm using "accompanied" because the worked "supported" may be overloaded in math.  I'm looking for some compact synonym of "independent and complementary".  The main thing is to demonstrate through examples that the relationship is many-to-many.
 A: Regarding your first question, I am assuming you meant to use the space $[0,1]$ rather than the whole set of reals (otherwise, it would not be a probability space). Besides that, or the most part, it does not matter. More precisely, given any real-valued random variable $X$, you can find a random variable $X'\colon [0,1]\to \mathbf R$ with the same distribution.
The same is true for random variables with values in any standard Lebesgue space, and in particular, any separable metric space. This implies that given any sequence $(X_n)_n$ of random variables $\Omega\to \mathbf R$, you can find a sequence $(X_n')_n$ of random variables $[0,1]\to \mathbf R$ with the same joint distribution.
On the other hand, it is not hard to see that there is no sequence $(X_\alpha)_{\alpha<\mathfrak c^+}$ of nontrivially i.i.d. random variables $[0,1]\to \mathbf R$. It should probably not be too hard to argue that there is no such uncountable sequence, even much shorter than $\mathfrak c^+$. So restricting the domain of the random variables does restrict the things we can see.
Since the structure of the domain (as opposed to the join distribution of variables) is usually mostly immaterial in probability theory, it is usually more convenient to leave the domain unspecified and implicit.
Regarding your second question, if there is a "the" standard probability space, then it would either be $[0,1]$ with the Lebesgue measure or $\{0,1\}^{\mathbf N}$ with the usual Haar/coin toss measure. Still, usually, you would speak of "a" standard probability space.
I'm not sure whether I understand your third question. The basic notion is that of a measurable space. Using this, we can define the notion of a measurable function (= random variable), a probability space (= a measurable space with a probability measure), and using those two, we can define the probability distribution (= the pushforward of the probability measure via the random variable). So I would not call these notions independent.
A: First of all, a note on terminology: the (cumulative) distribution function of a random variable $X$ is usually defined as
$$F_X(x) = P(\{\omega\in\Omega: X(\omega)\leq x\}.$$
Note here the $\leq$ instead of $<$.
Now let's get to your questions.
Q1: $(\mathfrak{R}, \mathfrak{B}(\mathfrak{R}), \mu)$ is not a probability space, because $\mu(\mathfrak{R}) = \infty.$ Instead, what we usually take is
$$([0, 1], \mathfrak{B}([0, 1]), \mu),$$
where $\mu$ is Lebesgue measure restricted to $[0, 1]$. This space can underly any probability distribution on $\mathfrak{R}.$ Note first of all that the identity function $\omega\mapsto \omega$ itself is a real-valued random variable and that it has a uniform distribution on $[0, 1].$ If we now know two distribution functions $F_X$ and $F_Y,$ then
$$X = F^{-1}_X(\omega), \quad Y = F^{-1}_Y(\omega)$$
have distribution functions $F_X$ and $F_Y$ respectively. $F^{-1}_X$ here denotes the generalized inverse of $F_X.$ To see that this is true, see here. This means that this space indeed underlies $X$ and $Y$.
Q2: This space does not satisfy the definition of a standard probability space that you mention, since it is not complete. However, $(\mathfrak{R}, \mathfrak{B}(\mathfrak{R}), P_X)$ can be called a canonical space for the random variable $X$ in the context of stochastic processes. Here, $P_X$ is the distribution of $X$ (which is a measure on $\mathfrak{R}$). That is, $P_X((-\infty, a]) = F_X(a),$ which is enough to define $P_X$ on $\mathfrak{B}(\mathfrak{R}).$ Then the identity $\omega \mapsto \omega$ has distribution $F_X$ on this space. More generally, if you have a sequence of random variables $X_1, ..., X_n,$ the canonical probability space is $(\mathfrak{R}^n, \mathfrak{B}(\mathfrak{R}^n), P_X),$ where $P_X$ is the distribution of the vector $(X_1, ..., X_n),$ defined by
$$P_X((-\infty, a_1]\times ... \times (-\infty, a_n]) = P(X_1\leq a_1, ..., X_n\leq a_n).$$ Again, the identity then has the same distribution as the vector $(X_1, ..., X_n).$ So you can generealize this idea to a space for multiple random variables.
Q3: probability spaces and distributions are not independent, because as you note, we require probability spaces to be able to define distributions. That is, theoretically, we first construct a probability space $(\Omega, \mathcal{F}, P).$ Then we define a random variable $X: \Omega\to \mathfrak{R}$ and we can consider its distribution function $F_X(x) = P(\{\omega\in\Omega: X(\omega)\leq x\})$. That is, a distribution requires the existence of a probability space with a random variable. However, in practice, it suffices to only consider the distribution and forget about the underlying probability space, but this is not always the case, especially when you start getting into stochastic processes and you need to be a bit more careful about measurability concerns. Furthermore, note that a distribution is not associated to a particular probability space and random variable, it just requires that there exists one.
In practice, we usually forget about the fact that such a probability space needs to exist, because it turns out that for any potential distribution function $F:\mathfrak{R}\to [0,1]$ that is non-decreasing, right-continuous with $\lim_{x\to-\infty}F(x) = 0, \lim_{x\to\infty}F(x)=1$, there exists a probability space with a random variable such that it has cumulative distribution function $F.$ We have actually already seen this: the construction in Q1 works for any such $F.$ Hence, we can just dream up a function satisfying these requirements and we can be certain that there exists some probability space with a random variable with that function as its distribution function.
A: Some concepts/definitions that might help:
A probability measure on $\left(\mathbf{R}^d, \mathcal{B}(\mathbf{R}^d) \right)$ is called distribution. The triplet obtained can be called a distribution space to distinguish it from the general probability space.
Typical distributions are built from Lebesgue measure and  $\mathcal{B}(\mathbf{R}^d)$-measurable functions $h:\mathbf{R}^d\rightarrow [0,\infty) $  with
$$ \int_{\mathbf{R}^d} h(x) \mu(dx) =1$$
by
$$ P_h(B) = \int_B h(x) \mu(dx) $$
for all $B\in \mathcal{B}(\mathbf{R}^d)$.
An example of distribution that cannot be built this way is Dirac's distribution concentrated at some point $x_0 \in \mathbf{R}^d$:
$$ \delta_{x_0} (B) = 1_{x_0\in B}$$
for all $B\in \mathcal{B}(\mathbf{R}^d)$.
Also, given probability space $\left(\Omega, \mathcal{F}, P\right)$ and $X:\Omega\rightarrow \mathbf{R}^d$ which is
$\mathcal{F}/\mathcal{B}(\mathbf{R}^d)$-measurable, one can build a distribution $P_X$ as follows:
$$ P_X = P \circ X^{-1}, $$
usually called the distribution of $X$ (or law of $X$), which suggests that now one can focus only on the distribution space $\left(\mathbf{R}^d, \mathcal{B}(\mathbf{R}^d), P_X \right)$.
Note: If $\Omega = \mathbf{R}^d, \mathcal{F} = \mathcal{B}(\mathbf{R}^d)$ and $P$ is a distribution, then taking $X$ to be the identity function,  $id$, we have:
$$ P_{X} = P.$$
Note 2: Two random variables, possibly defined on different spaces, can have the same distribution (law).
If $X$ is defined on an abstract space $\left(\Omega, \mathcal{F}, P\right)$ as above, it induces distribution
$ P_X$.
Then random variable $id$ defined on $\left(\mathbf{R}^d, \mathcal{B}(\mathbf{R}^d), P_X \right)$ has the same distribution.
Many models rely on knowing the distribution
of a random variable  $X$ rather than its explicit form and the  probability space on which it is defined.
Note 3: To answer Q3, I guess, we have the following facts:

*

*A distribution space is just a particular case of probability space.


*Yes, for a distribution, be it $P_h$  or Dirac type, there is always a random variable on a 'supporting' probability space that induces the same distribution: we take the probability space to be the starting distribution space itself and the random variable to be the identity function.


*(Complementing Note 2) If $A,B\in \mathcal{F}$ are different events such that $P(A)=P(B)$, then
$$1_A \not= 1_B,$$ but they are random variables with the same distribution, that is
$$ P_{1_A} = P_{1_B}.$$


*If $\alpha: \left(\mathbf{R}^d, \mathcal{B}(\mathbf{R}^d)\right) \rightarrow \left(\mathbf{R}^f, \mathcal{B}(\mathbf{R}^f) \right) $ is measurable, then

$$ P_{\alpha \circ X} = P_X \circ \alpha^{-1}. $$
Note 4: I finally realized that you are focusing on the distribution function.
A function $F:\mathbf{R}\rightarrow \mathbf{R}$ which is non-decreasing, bounded, left-continuous and for which $$\lim_{x\rightarrow -\infty} F(x) = 0$$
is called a distribution function. This definition stands on its own (no mention of measures).
The following facts can be proven.
Fact: Let $F$ be a distribution function such that
$$\lim_{x\rightarrow \infty} F(x) = 1.$$
Let also $m$ be a measure on $\left((0,1), \mathcal{B}((0,1))\right)$
such that
$$ m((0,x))=x $$
for all $x\in (0,1]$ (its existence can be proven).
Then there is a non-decreasing function $f:(0,1) \rightarrow \mathbf{R}$ such that measure $m\circ f^{-1}$ has $F$ as distribution function, that is
$$ (m\circ f^{-1})((-\infty,x)) = F(x)$$
for all $x\in \mathbf{R}$.
Fact 2: A measure $\mu$ on $(\mathbf{R}, \mathcal{B}(\mathbf{R}))$ is perfectly determined by its distribution function $F_\mu$ defined as
$$ F_\mu(x) = \mu ((-\infty,x)) $$
for all $x\in \mathbf{R}$. That is, if two measures on $(\mathbf{R}, \mathcal{B}(\mathbf{R}))$ have the same distribution function, they coincide.
These suggests that specifying the triplet
$$\left(\mathbf{R}, \mathcal{B}(\mathbf{R}), m\circ f^{-1}\right)$$
for some non-decreasing $f$ or rather a distribution function $F$ (with $\lim_{x\rightarrow \infty} F(x) = 1$, for which we know such $f$ exists) is the essential step in setting up any distribution space.
For a random variable on an abstract probability space, $X:(\Omega, \mathcal{F}, P) \rightarrow  (\mathbf{R}, \mathcal{B}(\mathbf{R}))$, as soon as we get $P_X$, the associated distribution, and $F_X$ its distribution function, as defined in the book, we are done (can forget about $X$, in some sense; basically replace it with $id$ introduced in Note 2, as it has the same distribution). Note that:
$$ F_X = F_{P_X} $$
with the second term defined above (in Fact 2).
A: Since Q1 and Q2 are well answered by other answers, I would like add some more details about Q3. Hope I correctly grasped the point of your question.

Although the meaning of distribution slightly varies across the literature and is sometimes misused, we can give a satisfactory definition that works in any abstract setting.

Let $X : \Omega \to \mathcal{S}$ be a $\mathcal{S}$-valued random variable from the probability space $(\Omega, \mathcal{F}, P)$ to a measurable space $(\mathcal{S}, \Sigma)$. In other words, it is a measurable function from $(\Omega, \mathcal{F})$ to $(\mathcal{S}, \Sigma)$.1) Then $X$ induces a probability measure $\mu$ on $(\mathcal{S}, \Sigma)$ via2)
$$ \forall E \in \Sigma \ : \quad \mu(E) = P(X \in E) = P(X^{-1}(E)) = P(\{\omega\in\Omega : X(\omega) \in E\}). $$
Then this $\mu$ is called the distribution of $X$.

Example 1. Let $\Omega = \{-1, 0, 1, 2\}$ be equipped with the power-set $\sigma$-algebra $\mathcal{F}=2^{\Omega}$ and the normalized counting measure $P(E) = \frac{1}{4}\#E$. Then

*

*$X_1 : \Omega \to \mathbb{R}$ defined by $X_1(\omega) = \omega$ has the distribution $\mu_1$ on $\mathbb{R}$ given by
$$ \mu_1(E) = \frac{1}{4} \mathbf{1}_{\{-1 \in E\}} + \frac{1}{4} \mathbf{1}_{\{0 \in E\}} + \frac{1}{4} \mathbf{1}_{\{1 \in E\}} + \frac{1}{4} \mathbf{1}_{\{2 \in E\}} $$
for any Borel subset $E$ of $\mathbb{R}$.


*$X_2 : \Omega \to \mathbb{R}$ defined by $X_2(\omega) = \omega^2$ has the distribution $\mu_2$ on $\mathbb{R}$ given by
$$ \mu_2(E) = \frac{1}{4} \mathbf{1}_{\{0 \in E\}} + \frac{1}{2} \mathbf{1}_{\{1 \in E\}} + \frac{1}{4} \mathbf{1}_{\{4 \in E\}} $$
for any Borel subset $E$ of $\mathbb{R}$.


*$X_3 : \Omega \to \{0,1,4\}$ defined by $X_3(\omega) = \omega^2$ has the distribution $\mu_3$ on $\mathcal{S}=\{0,1,4\}$ given by
$$ \mu_3(E) = \frac{1}{4} \mathbf{1}_{\{0 \in E\}} + \frac{1}{2} \mathbf{1}_{\{1 \in E\}} + \frac{1}{4} \mathbf{1}_{\{4 \in E\}} $$
for any subset $E$ of $\mathcal{S}$.3)
Example 2. Let $\Omega=[0,1]^2$ be equipped with the probability measure $P$ which is the Lebesgue measure restricted onto $[0, 1]^2$. Then

*

*$X_4 : \Omega \to \mathbb{R}$ defined by
$$ X_4(\omega_1, \omega_2) = \begin{cases}
0, & \text{if } \omega_1 \in [0, \frac{1}{4}); \\
1, & \text{if } \omega_1 \in [\frac{1}{4}, \frac{3}{4}); \\
4, & \text{if } \omega_1 \in [\frac{3}{4}, 1); \\
2020, & \text{if } \omega_1 = 1;
\end{cases} $$
has the same distribution as $X_2$.


*$X_5, X_6 : \Omega \to \mathbb{R}$ defined by
$$ X_5(\omega_1, \omega_2) = \begin{cases}
-\log \omega_1, & \text{if } \omega_1 \in (0, 1]; \\
42, & \text{if } \omega_1 = 0;
\end{cases} \qquad
X_6(\omega_1, \omega_2) = \begin{cases}
-\log (1-\omega_2), & \text{if } \omega_2 \in [0, 1); \\
1, & \text{if } \omega_2 = 1;
\end{cases} $$
have the same distribution, which is the exponential distribution of unit rate. In other words, they induce the same probability measure $\mu_{5}$ on $\mathbb{R}$ defined by
$$\mu_{5}(E) = \int_{E} e^{-x} \mathbf{1}_{(0,\infty)}(x) \, \mathrm{d}x $$
for any Borel subset $E$ of $\mathbb{R}$.
The information about $\mu_5$ may be encoded in a different way using the cumulative distribution function (CDF). The CDF $F_{X_5}$ of $X_5$ is given by
$$ F_{X_5}(x) = P(X_5 \leq x) = \mu_5((-\infty, x]) = \begin{cases}
0, & \text{if } x < 0; \\
1 - e^{-x}, & \text{if} x \geq 0;
\end{cases} $$
Of course, we have $F_{X_5} = F_{X_6}$ in this example.


*Define $X_7 : \Omega \to \mathbb{R}^2$ by $X_7(\omega) = (X_5(\omega), X_6(\omega))$. Then its distribution $\mu_7$ is given by
$$ \mu_7(E) = \iint_{E} e^{-x-y}\mathbf{1}_{(0,\infty)^2}(x,y) \, \mathrm{d}x\mathrm{d}y $$
for any Borel subset $E$ of $\mathbb{R}^2$. It turns out that $\mu_7 = \mu_5 \otimes \mu_5$ is the product of two copies of $\mu_5$, and its probabilistic implication is that $X_5$ and $X_6$ are independent.
Example 3. Let $\mu$ be any probability distribution on $\mathbb{R}$, and let $(\Omega, \mathcal{F}, P) = (\mathbb{R}, \mathcal{B}(\mathbb{R}), \mu)$. Also define $X_8(\omega) = \omega$. Then $X_8$ has the distribution $\mu$. For this reason, we often consider the notion of distribution without explicit reference to a random variable. For example, the standard normal distribution is the probability measure on $\mathbb{R}$ defined by
$$ E \mapsto \int_{E} \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \, \mathrm{d}x $$
for any Borel subset $E$ of $\mathbb{R}$. In this regard, we may as well say that the word distribution also stands for the honorable title given to a well-studied probability measure on a familiar space.
This construction also tells that, as long as we are only interested in dealing with a single random variable, the abstract notion of probability spaces is rather redundant and we can stick to this particular realization on $\mathbb{R}$. However, such notion provides great flexibility in developing various concepts under a unified framework and allowing to deal with them systematically.

1) If the term 'measurable space' is not familiar to you, you may regard $(\mathcal{S}, \Sigma)$ as the Euclidean space $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$ equipped with the Borel $\sigma$-algebra. Also, you do not worry too much about what it means by a measurable map at this point.
2) For this reason, $\mu$ is sometimes called the pushforward of $P$ by $X$ and denoted by $\mu = P \circ X^{-1}$.
3) Technically speaking, $\mu_2$ and $\mu_3$ are different distributions. However, they convey the same amount of information, and so, such difference will never affect any conclusion about the 'randomness' of $X_2$ or $X_3$. My personal impression is that the choice $X_3$ seems to be preferred in elementary probability textbooks for its simplicity, whereas $X_2$ is a more common choice in the literature because this allows to compare different distributions systematically.
