Let \((X, \mathscr{A})=(\mathbb{R}, \mathscr{B}(\mathbb{R}))\) and let \(\lambda\) be one-dimensional Lebesgue measure.

(i) A point \(x\) with \(\mu\{x\}>0\) is an atom. Show that every measure \(\mu\) on \((\mathbb{R}, \mathscr{B}(\mathbb{R}))\) which has no atoms can be written as image measure of \(\lambda\).

[\(\mu\) has no atoms, so \(F_{\mu}\) is continuous. Thus, \(G=F^{-1}\) exists and can be made left-continuous. Finally \(\mu[a, b)=F_{\mu}(b)-F_{\mu}(a)=\lambda\left(G^{-1}\{[a, b)\}ight)\). \(]\)

(ii) Is (i) true for measures with atoms, say, \(\mu=\delta_{0}\) ?

[ determine \(F_{\delta_{0}}^{-1}\). Is it measurable?]