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Let $\mathbb{P} : \mathcal{B} \to [0,1]$ be a real probability measure, where $\mathcal{B}$ is the Borel $\sigma-$algebra. Given the specific form of $\mathbb{P}$, we are asked to show that $\mathbb{P}$ is (absolutely) continuous. The way I'd proceed is to calculate the probability of a sub-interval $[a,b]$, writing it in terms of the integral of a density $f:\mathbb{R}\to [0,\infty)$. However, sub-intervals are not the whole Borel $\sigma-$algebra, so shouldn't I do the same calculations as above for any Borel set? Is there some theorem that states that calculating the probability on sub-intervals is enough to characterize the probability uniquely?

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  • $\begingroup$ If $\mathbb P (A)=\int_A f(x)dx$ for every interval $A=[a,b]$ (where $f$ is a non-negative measurable function) then the same holds for every Borel stet $A$. $\endgroup$ Apr 7, 2023 at 11:18

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The theorem you are looking for is the Carathéodory Extension theorem. If two probability measures agree on the set of intervals in $\mathbb{R}$, then they agree on the $\sigma$-algebra generated by the intervals, which is $\mathcal{B}$.

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  • $\begingroup$ Carathéodory does ring a bell. Thanks a lot. $\endgroup$ Apr 7, 2023 at 11:29

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