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