any $\sigma$-finite measure is equivalent to a probability measure

The following theorem states that for any $\sigma$-finite (http://planetmath.org/SigmaFinite) 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.

Theorem.

Any nonzero $\sigma$-finite measure $\mu$ on a measurable space $\mathrm{(}X\mathrm{,}\mathrm{A}\mathrm{)}$ is equivalent to a probability measure $\mathrm{P}$ on $\mathrm{(}X\mathrm{,}\mathrm{A}\mathrm{)}$. In particular, there is a positive measurable function $f\mathrm{:}X\mathrm{\to }\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{)}$ satisfying $\mathrm{\int }f\mathit{}𝑑\mu \mathrm{=}\mathrm{1}$, and $\mathrm{P}\mathit{}\mathrm{(}A\mathrm{)}\mathrm{=}{\mathrm{\int }}_{A}f\mathit{}𝑑\mu$ for all $A\mathrm{\in }\mathrm{A}$.

Proof.

Let $A_1,A_2,\mathrm{\dots }$ be a sequence in $𝒜$ such that and ${\bigcup }_k{A}_k=X$. Then it is easily verified that

$$g\equiv \sum _{k=1}^{\mathrm{\infty }}{2}^{-k}\frac{1_{A_k}}{1+\mu (A_k)}$$

satisfies $1\ge g>0$ and . So, setting $f=g/\int g𝑑\mu$, we have $\int f𝑑\mu =1$ and therefore $\mathbb{P}(A)\equiv {\int }_{A}f𝑑\mu$ is a probability measure equivalent to $\mu$.∎