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Let $\mathcal{F}_0(\mathbb{R})$ be the field of finite disjoint union of right semi-closed intervals. Let $F:\mathbb{R}\rightarrow \mathbb{R}$ be a non-decreasing right continuous function.

Then the set function $\mu(a,b]=F(b)-F(a)$ on right semi-closed intervals is finitely additive on $\mathcal{F}_0(\mathbb{R})$. Lemma 1.4.3 in Robert Ash proves the countable additivity of $\mu$.

In the proof, $F_n(x)$ is defined as follow, $$ F_n(x) = \left\{ \begin{array}{ll} F(x) & \quad -n \leq x < n \\ F(n) & \quad n \leq x \\ F(-n) & \quad x < -n \end{array} \right. $$

Now $F_n(x)$ is assumed to be finite for all $n$. But this is not true for all distributions functions. Consider for e.g. $$ F(x) = \left\{ \begin{array}{ll} \frac{1}{-x+1} & x < 1 \\ \infty & \quad 1 \geq x \end{array} \right. $$

Please point out, where I am wrong.

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In Definition 1.4.1 of that book a distribution function is defined to be a map $F:\mathbb{R}\to\mathbb{R}$, so your $F$ is not a disitribution function in this sense. – Stefan Hansen Jan 19 '13 at 13:33
up vote 1 down vote accepted

According to your first paragraph, you require $F:\mathbb R \to \mathbb R$. Your $F$ in the example does not satisfy this condition.

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