Let $F:\mathbb R\to \mathbb R$ an increasing function. Why is there a unique measure $\mu$ s.t. $$\mu((a,b])=F(b)-F(a) \ \ ?$$

Indeed, I could imagine an other measure $\nu$ s.t. $$\nu((a,b])=\mu((a,b]),$$ for all $a,b$ s.t. $a<b$, but for a more exotic measurable set $A$, we could have $\mu(A)\neq \nu(A)$.

  • $\begingroup$ Is $\mu$ a measure defined on the Borel $\sigma$-algebra of $\Bbb R$? Depending on the $\sigma$-algebra you are working on, it is possible that $]a,b]$ is not measurable. $\endgroup$ May 31 '19 at 15:22

It is not that there is a unique measure, but rather that there is a unique Borel measure. This can be proved directly using Caratheodory's extension theorem -- basically, sticking to the Borel $\sigma$-algebra rules out the more complicated measurable sets you might be imagining.

(It should maybe be also mentioned that $F$ also needs to be right-continuous to generate the pre-measure that Caratheodory turns into a Borel measure. This wasn't stated in your question but is a typical assumption, especially in the context of probability and turning CDFs into measures.)

For more info, you might look up Stieltjes' integral -- the $\mu$ you mention is the measure that appears when integrating against a CDF in that context.


This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .