Let \(\mathcal B\) be the Borel sigma-algebra on \(\mathbb R\), and let \(\mu\) be a finite Borel measure.
If \(F_1\supseteq F_2\supseteq\cdots\) are Borel sets, prove \[ \mu(F_k)\longrightarrow \mu\!\left(\bigcap_{k=1}^\infty F_k\right). \]
Assume \(m(E)=0\) implies \(\mu(E)=0\) for every Borel set \(E\). Prove that for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[ m(E)<\delta\quad\Longrightarrow\quad \mu(E)<\varepsilon \] for every Borel set \(E\).
Solution
1. Continuity from above.
Set \[ F:=\bigcap_{k=1}^\infty F_k, \qquad E_k:=F_k\setminus F_{k+1}. \] Then the \(E_k\) are pairwise disjoint and \[ F_1=F\,\dot\cup\,\bigcup_{k=1}^\infty E_k. \] Since \(\mu(F_1)<\infty\), \[ \mu(F_1)=\mu(F)+\sum_{k=1}^\infty\mu(E_k). \] Also \[ F_n=F\,\dot\cup\,\bigcup_{k=n}^\infty E_k, \] so \[ \mu(F_n)=\mu(F)+\sum_{k=n}^\infty\mu(E_k). \] The tail of the convergent series tends to \(0\), hence \[ \mu(F_n)\to\mu(F). \]
2. Absolute continuity in the epsilon–delta sense.
Assume the conclusion fails. Then there exist \(\varepsilon_0>0\) and Borel sets \(E_n\) such that \[ m(E_n)<2^{-n} \qquad\text{and}\qquad \mu(E_n)\ge\varepsilon_0. \] Define \[ F_N:=\bigcup_{n\ge N}E_n, \qquad E:=\bigcap_{N=1}^\infty F_N=\limsup_{n\to\infty}E_n. \] Then \[ m(F_N)\le\sum_{n\ge N}2^{-n}\to0, \] so \[ m(E)=0. \] By hypothesis, \[ \mu(E)=0. \] On the other hand, the sets \(F_N\) decrease to \(E\), and \(\mu\) is finite, so part 1 gives \[ \mu(E)=\lim_{N\to\infty}\mu(F_N). \] But \(E_N\subseteq F_N\), hence \[ \mu(F_N)\ge\mu(E_N)\ge\varepsilon_0 \] for every \(N\). Therefore \[ \mu(E)\ge\varepsilon_0, \] a contradiction. Thus the desired \(\delta\) exists.