Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
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 '10 at 16:45

2 Answers 2

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 $2n\int_{-1/n}^{1/n} f(t) dt$.

share|improve this answer

one could be indeed misled by noting that on one hand $l(f)=f(0)$ if writen like $\int f(t) \mu(dt)$ then $\mu(dt)$ would not be absolutely continuous (AC) wrt the Lebesgue measure $\lambda$ ; where we took $f$ in $C_{0}$ (continuous) ; in this cas one could think \mu as defined by $\mu({0})=1$ and $\mu(B) =0$ if $B$ does not contain $0$ ; and hence $\mu$ is indeed not absolutely continuous wrt to Lebesgue measure ; But (see Yosida P 118 on the dual of $L^{\infty})$ : $\mu$ is simply defined by $\mu(A)= l(1_{A})$ ; and the fact that $\mu$ should be AC wrt Lebesgue follows simply from : $l(f) <= ||l|| \times |f|_{\infty}$ which you apply to $f=1_{A} : |1_{A}|_{\infty} = \lambda(A)$ ; and this is satisfied by the expression of $l$ : $l(f)= \lim 2n \int_{-1/n}^{1/n}f(t)dt$ ; hence $l$ is indeed AC wrt Lebesgue measure. Note on the other hand that $\mu(\{0\})=\lim 2n \int_{-1/n}^{1/n}1_{\{0\}}dt=0$ ; Younes Adlay

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.