If two measures agree on generating sets, do they agree on all measurable sets?

Solution 1:

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 $B \in \mathcal{B} (\mathbb{R})$.

Solution 2:

With your help, this is the proof that I have come up with if anyone reading this at a later time is interested.

My Solution :

First, suppose that $m(\mathbb R) = n(\mathbb R) < \infty$. We show that $m=n$ on $\mathcal B$. Consider the collection $\mathcal G = \{B \in \mathcal B: m(B) = n(B)\}$. We observe that $\mathcal G$ is a monotone class:

Suppose $A_i \uparrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$ and define $B_1 = A_1$, $B_2 = A_2 - A_1$, $\dots$, $B_n = A_n - \cup_{i=1}^{n-1}A_i$ so that $\cup_{i=1}^\infty A_i = \cup_{i=1}^\infty B_i$ where $\{B_i\}$ are pairwise disjoint. Notice that for $C,D \in \mathcal G$ with $C \subset D$, we have $$m(D-C) = m(D) - m(C) = n(D) - n(C) = n(D-C),$$ since $m$ and $n$ are finite measures. So $B_i \in \mathcal G$ for all $i \in \mathbb N$. Then $A \in \mathcal G$, since $$m(A) = m(\cup_{i=1}^\infty B_i) = \sum_{i=1}^\infty m(B_i) = \sum_{i=1}^\infty n(B_i) = n(\cup_{i=1}^\infty B_i) = n(A)$$

Now suppose that $A_i \downarrow A$ where $A_i \in \mathcal G$ for all $i \in \mathbb N$. Then, $$m(A) = \lim_{n\to\infty}m(A_n) = \lim_{n\to\infty}n(A_n) = n(A)$$ since $m$ and $n$ are finite measures. So $A \in \mathcal G$, which makes $\mathcal G$ a monotone class. More specifically, it makes $\mathcal G$ a monotone class that contains $\mathcal C = \{(a,b):a,b\in\mathbb R\} \cup \{\mathbb R,\emptyset\}$. Thus, $\mathcal M(\mathcal C) \subseteq \mathcal G$ where $\mathcal M(\mathcal C)$ represents the smallest monotone class containing $\mathcal C$.

Since $\mathcal C$ is a collection of subsets of $\mathbb R$ closed under finite intersections (any intersection is either an open interval or the emptyset) containing $\mathbb R$, we can apply the Monotone Class Theorem. Thus, we get that $\mathcal M(\mathcal C) = \sigma(\mathcal C) = \mathcal B$, and more importantly, $\mathcal B = \mathcal M(\mathcal C) \subseteq \mathcal G$. By definition $\mathcal G \subseteq \mathcal B$, so we have $\mathcal G = \mathcal B$ and $m = n$ on $\mathcal B$.

Now suppose that one of the measures is not finite and consider the following increasing sequence of sets $A_1 = (-1,1)$, $A_2 = (-2,2)$, $\dots$, $A_k = (-k,k)$. Define $m_k(B) = m(B\cap A_k)$ and $n_k(B) = n(B \cap A_k)$ for all $B \in \mathcal B$. Then, $m_k, n_k$ are measures for every $k \in \mathbb N$, and moreover, are finite by monotonicity of measures (they are subsets of sets with finite measure). Therefore, since $m_k(\mathbb R) = n_k(\mathbb R) < \infty$ for all, we can apply the finite case as proved above to get that $m_k = n_k$ on $\mathcal B$ for all $k \in \mathbb N$. Then for any $B \in \mathcal B$, we have $$m(B) = \lim_{k\to\infty}m(B \cap (-k,k)) = \lim_{k\to\infty}m_k(B) = \lim_{k\to\infty}n_k(B) = \lim_{k\to\infty}n(B\cap (-k,k)) = n(B).$$ Thus, $m=n$ on $\mathcal B$.

Solution 3:

Lemma 7.1.2. (p. 68) of Measure Theory, volume 1, Vladimir I. Bogachev:
If two finite signed Borel measures on any topological space coincide on all open sets, they coincide on all Borel sets.

Its simple proof uses:
Lemma 1.9.4. If two probability measures on a measurable space $(X,A)$ coincide on some class $E\subset A$ that is closed with respect to finite intersections, then they coincide on the $\sigma$-algebra generated by $E$.

Link to Lemma 7.1.2 A related question: If two Borel measures coincide on all open sets, are they equal?

Solution 4:

The Borel sigma algebra on $\mathbb R$ has various equivalent generating sets: obviously the open intervals $(a, c)$ and also the half open intervals $[b, c)$ . The half open intervals are a semi-ring, and by Carathéodory a sigma finite measure (which you have here since each (finite) open interval has finite measure) defined on a semi-ring extends uniquely to a measure on the sigma-algebra generated from it.

So, if $m = n$ for each half open interval you are done.

For any half open interval $[b, c)$ there is $a \in \mathbb R $ with $a < b$.
Then $[b, c) = (a, c)\setminus (a, b)$ and $(a, b)\subset (a, c)$
So for any measure $\mu$ where $\mu( (a, b) ) $ is finite $\mu( [b, c)) = \mu((a, c)) - \mu((a, b))$
And since $m = n$ for each open interval they therefore agree on the half open intervals. $\blacksquare$