1

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?

0xbadf00d
  • 13,422
  • Is $E$ finite dimensional or could it also be infinite dimensional? I think that could change a lot the difficulty of the problem – Lucio Feb 15 '18 at 19:39
  • @Lucio No additional assumptions on $E$. It can be infinite-dimensional. – 0xbadf00d Feb 15 '18 at 19:46
  • Anyway, in the case $E=\mathbb{R}$, you can show 2) implies 1) by means of Radon-Nikodym theorem https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem and aboslute continuity of Lebesgue integral https://math.stackexchange.com/questions/535185/absolute-continuity-of-the-lebesgue-integral – Lucio Feb 15 '18 at 19:57
  • In the case $E$ is finite dimensional, wlog by choosing a basis you can work with $E=\mathbb{R}^n$ and in that case $\mu$ will be of the form $\mu=(\mu_1,\ldots, \mu_n)$ with $\mu_i$ real valued and $\mu_i <<\nu$ so you can reduce to the previous case – Lucio Feb 15 '18 at 19:58
  • The problem with infinite dimensional $E$ is that in general "good" basis do not exist and Radon-Nikodym theorem does not hold, so none of the above resonings work – Lucio Feb 15 '18 at 20:00

1 Answers1

2

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).

Martin Argerami
  • 205,756
  • Note that there is no need to explicitly assume boundedness of $\mu$: Since $\mu$ is $\sigma$-additive on a $\sigma$-algebra, it is strongly additive (i.e. $\left(\mu(S_n)\right){n\in\mathbb N}$ is summable for all disjoint $(S_n){n\in\mathbb N}\subseteq\mathcal S$) and hence bounded (i.e. $\sup_{S\in\mathcal S}\left|\mu(S)\right|_E<\infty$). (Don't sure what you mean by "uniformly" bounded, since $\mu$ doesn't depend on more than one parameter.) – 0xbadf00d Feb 15 '18 at 23:28
  • Yes, you are right. – Martin Argerami Feb 15 '18 at 23:31