While reading Robert Ash's Basic Probability Theory, I saw the following theorem (p.53):

Let $f$ be a nonnegative real-valued function on $\mathbb{R}$, with $\int_{-\infty}^\infty f(x)\,dx=1$. There is a unique probability measure $P$ on the Borel subsets of $\mathbb{R}$, such that $P(I)=\int_If(x)\,dx$ for all intervals $I$ in $\mathbb{R}$.

My question is this: If there is no such density function $f(x)$, then is it possible that a probability measure on Borel subsets of $\mathbb{R}$ is not completely determined by its values on intervals?

  • $\begingroup$ Since the Borel subsets are generated by the intervals, it is sufficient to specify them on intervals. $\endgroup$
    – copper.hat
    Sep 5, 2020 at 6:08

1 Answer 1


Finite disjoint unions of ingtervals of the type $[a,b)$ form an algebra which generates the Borel sigma algebra. This implies that if two probability measures agree on intervals they agree on all Borel sets.

  • $\begingroup$ Could you elaborate on the statement "Finite disjoint unions of intervals of the type [a,b) form an algebra which generates the Borel sigma algebra" ? Thank you. $\endgroup$
    – ashpool
    Sep 7, 2020 at 8:21
  • $\begingroup$ That is quite standard. A good reference for that is Section 8$ in Halmos's 'Measure Theory'. @ashpool $\endgroup$ Sep 7, 2020 at 8:30
  • $\begingroup$ Thanks. I read it. But the section you referred to makes no mention of Borel sigma algebra. Are you saying that every Borel set is a countable union of disjoint half-open intervals? This post says otherwise: math.stackexchange.com/questions/3462061/… $\endgroup$
    – ashpool
    Sep 7, 2020 at 8:50
  • $\begingroup$ Sigma algebra generated by a family of sets is an abstract concept;you cannot write down all sets in it explicitly in terms of the original family of sets. I am only saying that the sigma algebra generated by half closed intervals coincides with the Borel sigma algebra but nobody knows how a general Borel set looks like. $\endgroup$ Sep 7, 2020 at 8:55
  • $\begingroup$ If the half-open intervals are not disjoint, then one can't use the countable additivity of probability measure, so I don't see why two probability measures agreeing on intervals agree on every Borel set... $\endgroup$
    – ashpool
    Sep 7, 2020 at 8:58

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