# Are $\sigma$-finite measures agreeing on a generating set equal?

Suppose I have two measures $$m,n$$ and that they are both $$\sigma$$-finite and agree on some semi-ring $$R$$ generating their $$\sigma$$-algebras. Must these two measures then agree on the entire $$\sigma$$-algebra?

Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $$m,n$$ to $$R$$ are themselves $$\sigma$$-finite, then their extension to the $$\sigma$$-algebra is unique, but I suspect there might be a case where the restriction of a $$\sigma$$-finite measure to the semi-ring might not remain $$\sigma$$-finite.

• Let $\mathcal{B}$ be the $\sigma$-algebra. Then $\{A \in \mathcal{B} : m(A) = n(A)\}$ is a sigma-algebra containing $R$ and is thus $\mathcal{B}$. – mathworker21 Dec 1 '18 at 13:58
• Don't we run into a problem because the measures may not agree on complements of sets of infinite measure? – Bar Alon Dec 1 '18 at 14:23

## 1 Answer

It turns out that the answer is no.

Consider the two measures $$m,n$$ defined on the borel $$\sigma$$-algebra $$\mathcal{B}(\mathbb{R})$$ as follows: $$m(A)=|A\cap\mathbb{Q}|,\quad n(A)=|A\cap(\mathbb{Q}\cup\{\sqrt{2}\})|$$ Since $$\mathbb{Q}$$ is countable, they are both $$\sigma$$-finite. Additionally, they agree on the semi-ring $$R$$ of all half-open intervals (for any $$a,b\in \mathbb{R}$$: $$m([a,b))=\infty=n([a,b))$$), and $$R$$ indeed generates $$\mathcal{B}(\mathbb{R})$$.

On the other hand $$m(\{\sqrt{2}\})\not =n(\{\sqrt{2}\})$$ which is clearly a borel set.

Note that in the above example $$m, n$$ are indeed no longer $$\sigma$$-finite when restricted to $$R$$.