# Why is this probability measure countably additive?

I am trying to prove that the following probability measure given in my book (Rohtagi) is countably additive. My analysis is a bit rusty so if someone can explain to me the properties of integrals involved in proving that the probability measure is indeed countably additive I would be very grateful.

Let $(\Omega=(0,\infty),\mathbb{B})$ be a sample space. Here $\mathbb{B}$ is the Borel $\sigma$-field on $\Omega$. (The little bit of measure theory I remember tells me that a $\sigma$-field is a non-empty collection of subsets of $\Omega$ which is closed under countable union, complements and contains $\emptyset$. A Borel $\sigma$-field on $\Omega$ is the smallest $\sigma$-field on $\Omega$ containing all intervals). Let $P$ be defined for each interval $I$ as $PI = \int_I e^{-x}dx$. Now I want to prove that $P$ is countably additive. I am not sure how to do that for an arbitrary disjoint sequence of Borel sets.

Enumerate your countably many disjoint Borel sets as $B_n$ and observe that by finite additivity (which follows from considering the characteristic functions) we have $\int_{\cup_{n=1}^N B_n} e^{-x}dx = \sum_{n=1}^N \int_{B_n} e^{-x}dx$. Take limits on the left using the monotone convergence theorem (justified by the fact that $e^{-x}$ is positive) and on the right by the definition of an infinite sum to get the result.
• The sequence is $f_N(x) = e^{-x} \chi_{\cup_{n=1}^N B_n}(x)$. This sequence is pointwise nondecreasing because the exponential is always positive and its pointwise limit is $e^{-x} \chi_{\cup_{n=1}^\infty B_n}(x)$ – Chris Janjigian Feb 5 '12 at 7:19
• I think we should read $\int_{\bigcup_{n=1}^N B_n} e^{-x} dx = ...$ – Paul Pichaureau Feb 5 '12 at 12:14