From distribution to Measure

Question

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 ?

Answer 1

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?

Answer 2

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 $F$.