Let $X$ be the measurable functions with finite supremum norm on the unit interval. This is a banach space with respect to the supremum norm. The continuous functions on the unit interval have as dual the finite signed measures on the interval (theorem of Riesz). Is the same true for $X$ as well? I somehow doubt it, but what is the property that is special and only integration functionals have?

  • $\begingroup$ @david how does this answer my question? I am not considering $L^\infty$ $\endgroup$ – heinz Jul 17 '15 at 13:41
  • $\begingroup$ @heinz : What topology do you put on $\mathcal{L}^{\infty}$ then? It's necessay if you want to talk about the topological dual $\endgroup$ – Tryss Jul 17 '15 at 13:45
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    $\begingroup$ This is false tryss. I am not talking about the essential supremum seminorm, but about the norm you usually put on continous functions. It is a norm on bounded functions $\endgroup$ – heinz Jul 17 '15 at 14:46
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    $\begingroup$ @Heinz: Oh, so your $\mathcal{L}^{\infty}$ is not the usual $\mathcal{L}^{\infty}$... $\endgroup$ – Tryss Jul 17 '15 at 14:56
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    $\begingroup$ Related: The Duals of $l^\infty$ and $L^{\infty}$ $\endgroup$ – Martin Sleziak Jul 18 '15 at 5:11

To (i) maybe clarify your understandable confusion about finitely additive measures, and then to (ii) perhaps add a little more confusion and (iii) point out a different way of looking at the whole thing:

EDIT: Added a sketch of a proof that the dual is given by finitely additive measures. At the bottom...

I'm going to say $X$ is the space of bounded Borel functions on $[0,1]$. It's been said that $X^*$ is the space of finitely additive finite complex measures.

(i) First, say $\Lambda\in X^*$. Then define $$\mu(E)=\Lambda\chi_E.$$ It's clear that $\mu$ is finitely additive because $\Lambda$ is linear.

How do you define the integral? Well, of course it's a not Lebesgue integral because $\mu$ is not a measure. But it's clear what $\int\phi\,d\mu$ should be if $\phi$ is simple, and in fact for simple $\phi$ we have $$\Lambda\phi=\int\phi\,d\mu.$$Now if $f\in X$ there exist simple $\phi_n$ which tend to $f$ uniformly; now $\Lambda\phi_n\to\Lambda f$ since $\Lambda$ is bounded, while $\int\phi_n\,d\mu\to\int f\,d\mu$ if we define $\int f\,d\mu$ to be the limit.

(ii) That may clarify things. Now to add some confusion: A finitely additive measure is really nothing like what you think it is! At least it's nothing like what my mental picture was at first...

Example. Hahn-Banach says that there exists $\Lambda\in X^*$ such that if $\lim_{x\to0}f(x)$ exists then $$\Lambda f=\lim_{x\to0}f(x).$$Define $\mu$ as above.

You'd think that $\mu$ would be something like a point mass at the origin.

What is $\mu(\{0\})$? Look at the definitions: $\mu(\{0\})=0$. On the other hand, $\mu((0,1])=1$. Not a point mass at the origin at all...

Seems to me that if you're not confused you're not paying attention. What I take to be the moral here is you can't think of a finitely additive measure as distributing mass around the interval, as with an actual measure.

EDIT: The question was asked whether this $\mu$ is an actual countably additive measure. No, a measure can't act that way... Ah. Note that $\mu((0,1/n))=1$ for $n=1,2,\dots$. If $\mu$ were a measure it would follow that $\mu(\emptyset)=1$, hence that $\mu$ is not a measure.

(iii) Know anything about Banach algebras? The space $X$ is a Banach algebra; if $K$ is the maximal ideal space then it's not too hard to show that $X$ "is" $C(K)$ (that is, the Gelfand transform is an isometric isomorphism). If you look at it that way then $X^*$ becomes just the usual space of complex measures on $K$.

FREE BONUS: Turns out the theorem that the dual is what it is is not hard. When I first wrote this I hadn't figured out exactly what the definition of "finite" should be; the result really falls out of the definition.

(We take as done the combinatorial finagling needed to show the integral with respect to a finitely additive measure is well defined and linear on simple functions.)

So. Say a finitely additive measure $\mu$ is finite if there exists $c$ so that $$\sum_{j=1}^n|\mu(E_j)|\le c$$for any choice of finitely many disjoint sets $E_j$. Say $||\mu||$ is the smallest such $c$.

Now let $S$ be the set of simple functions. Since $S$ is dense in $X$, $X^*=S^*$.

It's easy to show that saying $\Lambda\phi=\int \phi\,d\mu$ gives a one-to-one correspondence between the (bounded or unbounded) linear functionals on $S$ and the (finite or not finite) finitely additive complex measures $\mu$. So we need to show that $\Lambda$ is bounded if and only if $\mu$ is finite. This is more or less clear once you notice that if $E_1,\dots,E_n$ are disjoint then there exist $a_j$ with $|a_j|=1$ such that $$\sum_{j=1}^n|\mu(E_j)|=\Lambda\left(\sum_{j=1}^na_j\chi_{E_j}\right).$$In fact it's clear that $||\mu||=||\Lambda||$.

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  • $\begingroup$ Thanks for your very instructive answer. In your example, is $\mu$ actually countably additive? For me its somewhat hard to tell. Otherwise I could use this to construct an actual example of a functional not induced by Integration wrt a countably additive measure. My problem is that i dont know a single non countably additive finitely additive measure on the unit interval. $\endgroup$ – heinz Jul 17 '15 at 20:37
  • $\begingroup$ @heinz Edited answer $\endgroup$ – David C. Ullrich Jul 17 '15 at 20:44
  • $\begingroup$ @heinz Bonus: added proof. Sort of $\endgroup$ – David C. Ullrich Jul 18 '15 at 3:16
  • $\begingroup$ Great! Thanks a lot $\endgroup$ – heinz Jul 18 '15 at 10:55

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