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

$$F(x)=\sum_{r_n<x}2^{-n}?$$ 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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I'm voting to close this question since what seems to be really asked is answered here and here – t.b. Mar 10 '12 at 8:56
The question is different, But I now that decomposition is closed in the strong topology! – checkmath Mar 12 '12 at 13:47
For the second part, your $F_k$ converge in total variation (hence also in weak-*) to the measure $\sum_{j=1}^\infty 2^{-j} \delta_{r_j}$, whose distribution function is the $F$ you give. To see this, note that the total variation of the tail $\sum_{j=k}^\infty 2^{-j} \delta_{r_j}$ is exactly $2^{-k+1}$ which goes to 0 as $k \to \infty$. But this measure is not absolutely continuous (and the function $F$ is not continuous). You might instead look at my suggestion here. – Nate Eldredge Mar 28 '12 at 14:47
Ok thanks, I see the measure is not absolutely continuous and it cannot be since converges in the strong topology! – checkmath Mar 28 '12 at 18:29