An element of $(L^{\infty})^*$ which does not seem to be a finitely additive abs. cont. measure. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:08:15Z http://mathoverflow.net/feeds/question/37730 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37730/an-element-of-l-infty-which-does-not-seem-to-be-a-finitely-additive-abs An element of $(L^{\infty})^*$ which does not seem to be a finitely additive abs. cont. measure. Dorian 2010-09-04T15:35:02Z 2010-09-04T16:43:13Z <p>Hi everyone, </p> <p>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 <em>absolutely continuous with respect to Lebesgue</em>. Therefore $l$ must be a finitely additive measure $&lt;&lt; dx$ on $[0,1]$.</p> <p>I apparently do not understand what this means for finitely additive measures since this element of <code>$(L^{\infty})^*$</code> 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?</p> <p>Best, Dorian</p> http://mathoverflow.net/questions/37730/an-element-of-l-infty-which-does-not-seem-to-be-a-finitely-additive-abs/37736#37736 Answer by Bill Johnson for An element of $(L^{\infty})^*$ which does not seem to be a finitely additive abs. cont. measure. Bill Johnson 2010-09-04T16:43:13Z 2010-09-04T16:43:13Z <p>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. $\lambda E &lt; \delta$ implies $\mu E &lt; \epsilon$. </p> <p>You can get a Hahn-Banach extension of $l$ by letting $l(f)$ be the limit through some free ultrafilter of $2n\int_{-1/n}^{1/n} f(t) dt$. </p>