# 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 $(X,\mathcal{A})$ is equivalent to a probability measure $\mathbb{P}$ on $(X,\mathcal{A})$. In particular, there is a positive measurable function $f\colon X\rightarrow(0,\infty)$ satisfying $\int f\,d\mu=1$, and $\mathbb{P}(A)=\int_{A}f\,d\mu$ for all $A\in\mathcal{A}$.

###### Proof.

Let $A_{1},A_{2},\ldots$ be a sequence in $\mathcal{A}$ such that $\mu(A_{k})<\infty$ and $\bigcup_{k}A_{k}=X$. Then it is easily verified that

 $g\equiv\sum_{k=1}^{\infty}2^{-k}\frac{1_{A_{k}}}{1+\mu(A_{k})}$

satisfies $1\geq g>0$ and $\int g\,d\mu<\infty$. So, setting $f=g/\int g\,d\mu$, we have $\int f\,d\mu=1$ and therefore $\mathbb{P}(A)\equiv\int_{A}f\,d\mu$ is a probability measure equivalent to $\mu$. ∎

Title any $\sigma$-finite measure is equivalent to a probability measure AnysigmafiniteMeasureIsEquivalentToAProbabilityMeasure 2013-03-22 18:33:44 2013-03-22 18:33:44 gel (22282) gel (22282) 6 gel (22282) Theorem msc 28A12 msc 28A10 SigmaFinite Measure