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This question is an exact duplicate of:

- From distribution function to probability measure

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- Prove that $\mu\left(\cup_{k=1}^\infty A_k\right)=\sum_{k\ge1}\mu(A_k)$

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 $F$.

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?

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