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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.

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  • $\begingroup$ 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 $\mathcal{B}$. $\endgroup$ – mathworker21 Dec 1 '18 at 13:58
  • $\begingroup$ Don't we run into a problem because the measures may not agree on complements of sets of infinite measure? $\endgroup$ – Bar Alon Dec 1 '18 at 14:23
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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 $R$.

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