8
$\begingroup$

Let $S$ be a countable discrete set, the following two results are quite easy to prove:

  1. Every countably additive probability measure $\mu$ on $S$ commutes (in Fubini's sense) with every finitely additive probability measure $\nu$ on $S$, i.e. $$ \int\int f(x,y)d\mu(x)d\nu(y)=\int\int f(x,y)d\nu(y)d\mu(x) $$ for all bounded $f$.

  2. The set of countably additive probability measures on $S$ is dense in the weak* topology in the set of all finitely additive probability measures on $S$.

My question is:

Question: Do those results hold also for uncountable sets?

More precisely, I am interested in the case when $S$ is a locally compact group and the measures are identified with positive functional on $L^\infty(S)$, and the latter is defined with respect to the Haar measure.

Update: Question 2., in its natural setting, as positive answer, as observed in the comments by Bill Johnson. Question 1. is still unanswered.

Thanks in advance,

Valerio

Improve this question
$\endgroup$
6
  • $\begingroup$ Isn't the answer to (2) trivially "yes" because $L^1(S)$ is weak$^*$ dense in its second dual by Banach-Alaoglu? $\endgroup$
    – Bill Johnson
    Dec 28, 2011 at 14:41
  • $\begingroup$ Maybe yes, even if it's not clear to me how you use Banach-Aloglu theorem, which concerns the compactness of the unit ball in the weak* topology, as far as I know. Also, since this should be a reference for a paper for game theorists, maybe I should write two lines to explain how any countably additive finite measure can be seen as an element of $L^1(S)$.. I think that in general it follows from the representation of the predual of the von Neumann algebra $L^\infty(S)$.. any more elementary argument? $\endgroup$
    – Valerio Capraro
    Dec 28, 2011 at 17:00
  • $\begingroup$ Sorry; I meant Goldstine's theorem. $\endgroup$
    – Bill Johnson
    Dec 28, 2011 at 17:43
  • 2
    $\begingroup$ Not every countably additive probability measure is given by an element of $L^1(S)$; just those that are absolutely continuous with respect to Haar measure. But this small class of measures is weak$^*$ dense in the finitely additive probability measures by Goldstine's theorem essentially. Formally you use Goldstine's theorem to approximate your finitely additive probability by an absolutely continuous countably additive signed measure of mass at most one, then take the positive part of the signed measure and normalize it--this will still be a good approximation if your weak$^*$ neighborhood $\endgroup$
    – Bill Johnson
    Dec 28, 2011 at 17:52
  • 1
    $\begingroup$ ...is given by integration against non negative $L^\infty(S)$ functions (which is WLOG). $\endgroup$
    – Bill Johnson
    Dec 28, 2011 at 17:53

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.