# Why is there a unique measure $\mu$ s.t. $\mu(a,b]=F(b)-F(a)$ where $F$ is increasing?

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. $$a, but for a more exotic measurable set $$A$$, we could have $$\mu(A)\neq \nu(A)$$.

• 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. 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.