# uniqueness of measures extended from a $\pi$-system

The following theorem allows measures to be uniquely defined by specifying their values on a $\pi$-system (http://planetmath.org/PiSystem) instead of having to specify the measure of every possible measurable set. For example, the collection of open intervals $(a,b)\subseteq\mathbb{R}$ forms a $\pi$-system generating the Borel $\sigma$-algebra (http://planetmath.org/BorelSigmaAlgebra) and consequently the Lebesgue measure $\mu$ is uniquely defined by the equality $\mu((a,b))=b-a$.

###### Theorem.

Let $\lambda$, $\mu$ be measures on a measurable space $(X,\mathcal{A})$. Suppose that $A$ is a $\pi$-system on $X$ generating $\mathcal{A}$ such that $\lambda=\mu$ on $A$ and that there exists a sequence $S_{n}\in A$ with $\bigcup_{n=1}^{\infty}S_{n}=X$ and $\lambda(S_{n})<\infty$. Then, $\lambda=\mu$.

###### Proof.

Choose any $T\in A$ such that $\lambda(T)<\infty$ and set $\mathcal{B}=\{S\in\mathcal{A}:\lambda(S\cap T)=\mu(S\cap T)\}$. For any $S\in A$, $S\cap T\in A$ and the requirement that $\lambda,\mu$ agree on $A$ gives $S\in\mathcal{B}$, so $\mathcal{B}$ contains $A$. We show that $\mathcal{B}$ is a Dynkin system in order to apply Dynkin’s lemma. It is clear that $X\in\mathcal{B}$. Suppose that $S_{1}\subseteq S_{2}$ are in $\mathcal{B}$. Then, the additivity of $\lambda$ and $\mu$ gives

 $\lambda\left((S_{2}\setminus S_{1})\cap T\right)=\lambda(S_{2}\cap T)-\lambda(% S_{1}\cap T)=\mu(S_{2}\cap T)-\mu(S_{1}\cap T)=\mu\left((S_{2}\setminus S_{1})% \cap T\right)$

and therefore $S_{2}\setminus S_{1}\in\mathcal{B}$. Now suppose that $S_{n}$ is an increasing sequence of sets in $\mathcal{B}$ increasing to $S\subseteq X$. Then, monotone convergence of $\lambda$ and $\mu$ gives

 $\lambda(S\cap T)=\lim_{n\rightarrow\infty}\lambda(S_{n}\cap T)=\lim_{n% \rightarrow\infty}\mu(S_{n}\cap T)=\lambda(S\cap T),$

so $S\in\mathcal{B}$ and $\mathcal{B}$ is a Dynkin system containing $A$. By Dynkin’s lemma this shows that $\mathcal{B}$ contains $\sigma(A)=\mathcal{A}$.

We have shown that $\lambda(S\cap T)=\mu(S\cap T)$ for any $S\in\mathcal{A}$ and $T\in A$ with $\lambda(T)<\infty$. In the particular case where $X\in A$ and $\lambda,\mu$ are finite measures then it follows that $\lambda(S)=\mu(S)$ simply by taking $T=X$. More generally, choose a sequence of sets $T_{n}\in A$ satisfying $\lambda(T_{n})<\infty$ and $\bigcup_{n}T_{n}=X$. For any $S\in\mathcal{A}$, $S_{n}\equiv(S\cap T_{n})\setminus\bigcup_{m=1}^{n-1}T_{m}$ is a pairwise disjoint sequence of sets in $\mathcal{A}$ with $S_{n}\subseteq T_{n}$ and $\bigcup_{n}S_{n}=S$. So, $\lambda(S_{n})=\mu(S_{n})$ and the countable additivity of $\lambda$ and $\mu$ gives

 $\lambda(S)=\sum_{n}\lambda(S_{n})=\sum_{n}\mu(S_{n})=\mu(S).$

Title uniqueness of measures extended from a $\pi$-system UniquenessOfMeasuresExtendedFromApisystem 2013-03-22 18:33:08 2013-03-22 18:33:08 gel (22282) gel (22282) 8 gel (22282) Theorem msc 28A12 LebesgueMeasure DynkinsLemma