Let
- $\Omega\subseteq2^\Omega$
- $\mathcal S\subseteq2^\Omega$
- $E$ be a normed $\mathbb R$-vector space
- $\mu:\mathcal S\to E$
- $\nu:\mathcal S\to[0,\infty)$
Write $\mu\ll\nu$, if $$\forall\varepsilon>0:\exists\delta>0:\forall S\in\mathcal S:\nu(S)<\delta\Rightarrow\left\|\mu(S)\right\|_E<\varepsilon\tag1\;.$$
I want to show that if $\mathcal S$ is a $\sigma$-algebra and $\mu$ and $\nu$ are $\sigma$-additive, then $(1)$ is equivalent to $$\forall S\in\mathcal S:\nu(S)=0\Rightarrow\mu(S)=0\;.\tag2$$
Clearly, $(1)$ implies $(2)$. How can we show the other implication?
This is the standard argument in Rudin's book.
If (1) is false, there exists $\varepsilon>0$ and sets $E_n\in\mathcal S$ with $\nu(E_n)<2^{-n}$ and $\|\mu(E_n)\|\geq\varepsilon$. Then $|\mu|(E_n)\geq\varepsilon$, where $|\mu|$ is the total variation of $\mu$: $$\tag{*} |\mu|(F)=\sup\{\sum\|\mu(F_n)\|:\ F=\bigcup_{n=1}^\infty F_n,\ \text{ disjoint}\}. $$ Of course, one needs to prove that (*) defines a measure, but this is also standard.
If $$B_n=\bigcup_{k\geq n}E_n,\ \ B=\bigcap_n B_n,$$ Then $\nu(B_n)<2^{-n+1}$ (after adding the geometric series) and $B_n\supset B_{n+1}$. By continuity of the measure, $\nu(B)=0$. Also $$ |\mu|(B)=\lim_n|\mu|(B_n)\geq\limsup_n|\mu|(E_n)\geq\varepsilon>0. $$ Hence (2) fails. The contradiction shows that (2) implies (1).