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Dears,

the question is - why there are finitely-additive measures in dual to L infinity, not only sigma-additive? I was thinking, that sigma-additive measures are related with embedded elements of L1 into (Linfinity)*, whereas finitely-additive [and not sigma-additive] are related with "the rest". Can someone explain that or give me link to article or pdf where it is well explained? I want to know what gives that difference.

robin3210
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  • σ -additivity corresponds of course precisely to weak∗-continuity of the functionals on L∞=(L1)∗) What does it mean precisely? – robin3210 Jun 04 '19 at 18:17

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I think the best think to do is to have an example of a functional in the dual of $L_\infty$ of some space (say the Borel space $\big(\mathbb{N}, 2^{\mathbb{N}}, \#\big)$ that is not a $\sigma$--finite measure. A classical example is to start with $\mathcal{c}$, the space of convergent sequences and define $\Lambda\phi:=\lim_n\phi(n)$. This is a continuous linear functional on $\mathcal{c}$. Using Hahn-Banach theorem, $\Lambda$ can be extended to the whole of $\ell_\infty$. Convince yourself that $\Lambda$ is not a $\sigma$--additive measure. It also follows from this that there is no $f\in\ell_1$ for which $\Lambda\phi=\int \phi f_1\,d\mu$ whenever $\phi\in\ell_\infty$.

In general, the finitely additive measures in a measurable space $(X,\mathscr{F})$ are go under the name of charges. There is also a notion of variation $|\nu|$ for a charge $\nu$. The charges in $L_\infty(X,\mathscr{F},\mu)$ also have the property that they are absolutely continuous with respect to the ambient measure $\mu$.

A modern and somewhat simple exposition of this things are in Aliprantis' Infinite dimensional Analysis, Hitchhiker's guide, chapter 14.

All this depends on the axiom of choice.

Mittens
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  • I understand all you have written; Every functional states the measure on every natural point (between 0 and 1 cause it has to be absolutely continous with respect to counting measure which gives every natural number a measure of 1). But if that functional gives the sequences its limes, then it should give measure 0 to every natural number (bcs in other case, its value would contain also the measure of some natural number n0 (1) multiplied by n0th element of sequence, which can be non-zero. – robin3210 Jun 04 '19 at 19:36
  • Hence, the measure [given by that functional] of every natural number is 0, however the integrate of any sequence over all N (which is the countable sum of natural numbers) is non-zero if that sequence is not convergent to 0. Am I right? – robin3210 Jun 04 '19 at 19:36
  • I just made some edits to give little more information and hopefully, to address your concern. – Mittens Jun 04 '19 at 20:01
  • One more question - measure in our space is sigma adiitive if and only if such f1 exists? – robin3210 Jun 04 '19 at 20:09
  • Of course! If a charge in $L^_\infty(X,\mathscr{F},\mu)$ happens to be $\sigma$--finite, then it is an absolutely continuous measure w.r.t $\mu$ and and then you can apply Radon-Nikodym. Again, $L^\infty$ is the space of all (finite) charges that are absolutely continuous w.r.t. $\mu$. A $\sigma$--finite charge is a what we usually call measure; hence, $L^*\infty(X,\mathscr{F},\mu)$ contains all finite measures in $(X,\mathscr{F})$ that are absolutely continuous w.r.t. $mu$. – Mittens Jun 04 '19 at 20:17