[Math] If two measures agree on generating sets, do they agree on all measurable sets

measure-theoryreal-analysis

Here's the problem that insprired my question:

Suppose $X$ is the set of real numbers, $\mathcal B$ is the Borel $\sigma$-algebra, and $m$ and $n$ are two measures on $(X, \mathcal B)$ such that $m((a,b)) = n((a,b)) < \infty$ whenever $-\infty < a< b < \infty$. Prove that $m(A) = n(A)$ whenever $A \in \mathcal B$.

When I look at this, I want to say that this problem is essentially trivial, but I can't convince myself that it actually is. I want to say that since these measures agree on sets that generate $\mathcal B$, and every $\mathcal B$-measurable set can be formed by taking countable intersections and unions of these types of sets, then the result follows from the fact that these are both measures.

So my questions are:

  1. Can we actually say that every Borel measurable set can be written as a countable union or intersection of these finite open intervals?

  2. Does the result immediately follow from properties of measures?

Best Answer

You can also look in to the Dynkin's $\pi - \lambda$ Theorem, which is equivalent with monotone class theorem.

Definition Let $P$ and $L$ be collections of subsets of a set $X$,

$P$ is a $\pi$-system if it is closed under finite intersections.

$L$ is a $\lambda$-system if the following hold:

  1. $\emptyset \in L$;
  2. if $A\in L$ then $A^c \in L$;
  3. $L$ is closed under countable disjoint unions.

Dynkin's $\pi - \lambda$ Theorem: Let $P$ be a $\pi$-system of subsets of $X$ and $L$ a $\lambda$-system of subsets of $X$. Suppose that $P\subset L$, then: $$\sigma(P) \subset L.$$

How to apply it:

  1. $P:=\{(a,b) : -\infty < a<b<\infty\}$ forms a $\pi$-system, and in general the collection of intervals (or boxes in higher dimensions) always forms a $\pi$-system.

  2. $L:=\{A\in \mathcal{B} (\mathbb{R}) : m(A) = n (A)\}$ forms a $\lambda$-system, this should be very easy to check. For now we only have that $L\subset \mathcal{B} (\mathbb{R})$.

  3. The assumption that $m((a,b)) = n((a,b))$ imples $P\subset L$

  4. By Dynkin's $\pi - \lambda$ Theorem, we know that $$\sigma(P) \subset L,$$ since $\sigma(P) = \mathcal{B} (\mathbb{R})$, thus $L = \mathcal{B} (\mathbb{R})$, which means the two measures agree on all $B \in \mathcal{B} (\mathbb{R})$.

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