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Consider $\mathcal{M}([0,1])\equiv C([0,1])^*$ (topological dual) then the Radon-Nikodym decomposition gives us

$\mathcal{M}([0,1])=AC([0,1])\oplus \mathcal{S}[0,1]$

Where $AC([0,1])$ denotes the space of measures that are absolutely continuous with respect to Lebesgue measure and $\mathcal{S}[0,1]$ are the singular ones.

Show that both spaces in this decomposition are not closed in the weak$^\ast$ topology.

To show that the first is not closed I chose a function $\varphi\geq0$ such that $\int_{0}^{1} \varphi=1$ and defined $\mu_j=j\varphi(x/j)$ for which it is well known that

$\mu_j\stackrel{*}{\rightharpoonup}\delta_0. $

For the second part it was a little bit more subtle I first chose $\{q_n\}$ an enumeration of the rationals on $[0,1]$ and let $\Lambda_n=\{q_1,q_2,...,q_n\}$ and defined

$F_k =\sum_{j=1}^{k}2^{-j}\delta_{r_j}$

Is there an easy way to prove that this sequence of singular measures converges Weakly$^\ast$ to

$F(x)=\sum_{r_n Notation: $\delta_a(f)=f(a)$

Can anyone indicate a text book where I can find this question and other questions like that?

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    Ok thanks, I see the measure is not absolutely continuous and it cannot be since converges in the strong topology!2012-03-28

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