The following theorem states that for any $\sigma$ -finite measure $\mu$ , there is an equivalent probability measure $\mathbb{P}$ -- that is, the sets $A$ satisfying $\mu(A)=0$ are the same as those satisfying $\mathbb{P}(A)=0$ . This result allows statements about probability measures to be generalized to arbitrary $\sigma$ -finite measures.

Proof. Let $A_1,A_2,\ldots$ be a sequence in $\mathcal{A}$ such that $\mu(A_k)<\infty$ and $\bigcup_kA_k=X$ . Then it is easily verified that \begin{equation*} g\equiv\sum_{k=1}^\infty 2^{-k}\frac{1_{A_k}}{1+\mu(A_k)} \end{equation*}satisfies $1\ge g>0$ and $\int g\,d\mu<\infty$ . So, setting $f=g/\int g\,d\mu$ , we have $\int f\,d\mu=1$ and therefore $\mathbb{P}(A)\equiv\int_Af\,d\mu$ is a probability measure equivalent to $\mu$ .