any $\sigma$-finite measure is equivalent to a probability measureAnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure2008-11-28 23:53:442008-12-14 13:44:08Theorem$\sigma$-finitesigma-finitemeasure% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{theorem*}{Theorem}
\PMlinkescapeword{equivalent}
The following theorem states that for any \PMlinkname{$\sigma$-finite}{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.
\begin{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 $\mathbb{P}(A)=\int_Af\,d\mu$ for all $A\in\mathcal{A}$.
\end{theorem*}
\begin{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 $\mu$.
\end{proof}