From distribution to Measure I have been asked to create a new post with my question. So it is about starting from a distribution function and proving that we can always find a probability space. My attempt is this : 
So assume we have a probability space $(\Omega, \mathcal{F}, P)$ and a random variable $X : \Omega \rightarrow \mathbb{R}^*$.
* Here we start with the distribution function $F$. increasing and right continuous, st. $F(\infty) = 1$ and $F(-\infty) = 0$. We define :
\begin{align}
P_I : I &\longmapsto [0,1] \quad I \in \mathbb{R}^2 \text{ (interval)} \\
[a,b] &\longmapsto F(b) - F(a)
\end{align}


*

*We can define $P_X$ on the borel sets by decomposing the set into disjoint intervals : 


\begin{align}
P_X : \mathcal{B}(\mathbb{R}^*) &\longmapsto [0,1] \quad B = \dot \bigcup B_i \quad B_i \in \mathbb{R}^2 \\
B &\longmapsto \sum_i P_I[a_i,b_i] 
\end{align}


*

*By taking $X$ as the identity function we have $\Omega = \mathbb{R}^*$ and $P = (P_X)_{\Omega}$


One user pointed out that not every borel sets is a countable union/ intersection of intervals. So my question is how to proceed from there to finish the demonstration ?
EDIT : 
I finally ended up with that :


*

*We can define $P_X$ on the algebra $\mathcal{A} = \{A:A = \sum_i [a_i, b_i]\}$ with $-\infty < a_i \leq b_i < \infty$ 


\begin{align}
P_X : \mathcal{A} &\longmapsto [0,1] \\
A &\longmapsto \sum_i P_I[a_i,b_i] 
\end{align}


*

*$P_X$ is a probability measure on the algebra $\mathcal{A}$ so it can be extended to the minimal $\sigma$-algebra containing all closed intervals. We have then $\sigma(\mathcal{A}) = \mathcal{B}(\mathbb{R})$. We just finally take this new measure $P$ and construct the probability space $(\mathbb{R}, \mathcal{B}(\mathbb{R}),P)$


Thoughts ?
 A: Let  $F:R\to [0,1]$ be a continuous from the right function on
$R$, which satisfies the following conditions: $$\lim_{x \to
-\infty}F(x)=0~\&~\lim_{x \to +\infty}F(x)=1.
$$
We suppose that $F(-\infty)=0$ and $F(+\infty)=1$.
We set   $\Omega=R \cup \{+\infty\}$.
Let ${\cal{A}}$ denote a class of all subsets of $\Omega$, which are
represented by the union of finite number of ”semi-closed from the right 
intervals” of the form $(a, b]$, i.e.,
$$
{\cal{A}}=\{ A|A=\sum_{i=1}^n(a_i,b_i]\},
$$
 where $-\infty \le a_i < b_i \le \infty (1 \le i \le n).$
It is easy to show that ${\cal{A}}$ is an algebra of subsets of
$\Omega$.
We set $$P(A) = P(\sum_{i=1}^n(a_i, b_i])) = \sum_{i=1}^n P((a_i,
b_i])=\sum_{i=1}^n F(b_i)-F(a_i).$$
One can easily demonstrate that the real-valued function $P$ is a
probability defined on ${\cal{A}}$ . Using Charatheodory well known theorem  about extension of the probability from the  algebra to the minimal sigma-algebra, we deduce
that there exists a unique probability measure $\overline{P}$ on
$\sigma({\cal{A}})$ which is an extension of $P$. Let $P_F$ denotes the restriction of the $\overline{P}$ to the $\sigma$-algebra $R \cap \sigma({\cal{A}})$.The class
$R \cap \sigma({\cal{A}})$ coincides with  Borel $\sigma$-algebra of
subsets of the real axis ${\bf R}$ which is denoted by ${\cal{B}}({\bf R})$. A real-valued function $P_F$ is called a probability Borel measure on ${\bf R}$ defined by the
distribution function $F$.
A: You easily defined $P_X$ on the compact set, so extend it as an inner regular measure: For any $A\in \mathcal B(\mathbb R^*)$,
$$P_X(A) = \sup \{ P_X(K) \, : \, \text{compact } K\subset A \}\,.$$
Is it good?
