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## An element of $(L^{\infty})^*$ which does not seem to be a finitely additive abs. cont. measure.

Hi everyone,

I have a question which I am quite stumped on. Consider the linear functional $l(f) = f(0)$ defined on $C([-1,1])$. By Hahn-Banach this linear functional can be extended to one on all of $L^{\infty}([-1,1])$. Now the space $(L^{\infty})^*$ is the set of all finitely additive measures which are absolutely continuous with respect to Lebesgue. Therefore $l$ must be a finitely additive measure $<< dx$ on $[0,1]$.

I apparently do not understand what this means for finitely additive measures since this element of $(L^{\infty})^*$ does not appear to be absolutely continuous; it is just dirac measure. Can someone help clarify this apparent inconsistency? Are the finitely additive functionals only defined on intervals $[a,b)$ or something of this nature?

Best, Dorian

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Well, the extension of $l$ is certainly not the dirac delta measure, because evaluating $L^\infty$ functions at a point doesn't make sense. Rather, it's some weird Hahn-Banach extension of a dirac delta measure. So, while I think this looks odd to start with, the more I think about it, the less I see a contradiction: remember that all your finitely additive measure has to do is integrate against a continuous function to evaluation at 0. Sorry, maybe that's not a very good answer, hence why it's just a comment. – Matthew Daws Sep 4 2010 at 16:45

The problem is with the concept of absolutely continuous FINITELY additive measure. Here AC just means that $\mu E = 0$ whenever the Lebesgue measure $\lambda E$ of $E$ is zero (that is, $\mu$ is a general finitely additive finite measure on the measure algebra generated by Lebesgue measure). The point is that this condition on $\mu$ does not imply that for every $\epsilon > 0$ there is $\delta > 0$ s.t. $\lambda E < \delta$ implies $\mu E < \epsilon$.
You can get a Hahn-Banach extension of $l$ by letting $l(f)$ be the limit through some free ultrafilter of $2n\int_{-1/n}^{1/n} f(t) dt$.