# This could totally be a stupid question but I'm unsure: is a measure (ie positive, countable additiv

This could totally be a stupid question but I'm unsure: is a measure (ie positive, countable additive on a $\sigma$ algebra, 0 for the empty set) actually a measurable function (wrt to the Borel-sigma algebra on $\mathbb{R}$)?
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esperoanow
For a measure space $\left(X,\mathcal{F},\mu \right)$ you have that
${\mu }^{-1}\left(\left(-\mathrm{\infty },c\right]\right)=\left\{A\in \mathcal{F}:\mu \left(A\right)⩽c\right\}$
so you will need a $\sigma$-algebra defined in $\mathcal{F}$ to define the measurability of $\mu$. Well, you can define this $\sigma$-algebra using $\mu$, this will give an induced $\sigma$-algebra in $\mathcal{F}$, and we can note it by $\sigma \left(\mu \right)$.

2d3vljtq
Given a function, a sigma algebra on the domain and a sigma algebra on the codomain, you can ask whether the function is measurable with respect to the given sigma algebras. You have only specified a sigma algebra on the codomain, so we can not answer your question.
What probably confused you is the fact that the domain of a measure is a sigma algebra on some set, so we need to consider a sigma algebra on another sigma algebra - the domain of the measure - to determine whether it is measurable.