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 $(a,b)\subseteq ℝ$ 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 $\mathrm{(}X\mathrm{,}\mathrm{A}\mathrm{)}$. Suppose that $A$ is a $\pi$-system on $X$ generating $\mathrm{A}$ such that $\lambda \mathrm{=}\mu$ on $A$ and that there exists a sequence $S_n\mathrm{\in }A$ with ${\mathrm{\bigcup }}_{n\mathrm{=}\mathrm{1}}^{\mathrm{\infty }}{S}_n\mathrm{=}X$ and . Then, $\lambda \mathrm{=}\mu$.

Proof.

Choose any $T\in A$ such that and set $ℬ=\{S\in 𝒜:\lambda (S\cap T)=\mu (S\cap T)\}$. For any $S\in A$, $S\cap T\in A$ and the requirement that $\lambda ,\mu$ agree on $A$ gives $S\in ℬ$, so $ℬ$ contains $A$. We show that $ℬ$ is a Dynkin system in order to apply Dynkin’s lemma.It is clear that $X\in ℬ$. Suppose that $S_1\subseteq S_2$ are in $ℬ$. 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 $S_2\setminus S_1\in ℬ$.Now suppose that $S_n$ is an increasing sequence of sets in $ℬ$ increasing to $S\subseteq X$. Then, monotone convergence of $\lambda$ and $\mu$ gives

$$\lambda (S\cap T)=\underset{n\to \mathrm{\infty }}{lim}\lambda (S_n\cap T)=\underset{n\to \mathrm{\infty }}{lim}\mu (S_n\cap T)=\lambda (S\cap T),$$

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

We have shown that $\lambda (S\cap T)=\mu (S\cap T)$ for any $S\in 𝒜$ and $T\in A$ with . 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 and ${\bigcup }_n{T}_n=X$. For any $S\in 𝒜$, $S_n\equiv (S\cap T_n)\setminus {\bigcup }_{m=1}^{n-1}{T}_m$ is a pairwise disjoint sequence of sets in $𝒜$ 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

$$\lambda (S)=\sum _n\lambda (S_n)=\sum _n\mu (S_n)=\mu (S).$$

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