Concentration set of a weak star limit of a sequence of measures. Let $I=[0,1]$, $\lambda$ the Lebesgue measure and $\mu$ a non negative Borel measure. Assume that $g_n$ is a sequence of non neagative continuous functions satisfying:
$$g_n(x)\to 0\ a.e.\ \mbox{and}\ \int_Ig_n(x)dm=1.$$
Note that $g_ndm$ are Borel measures. If $g_n dm$ converge in the weak star topology to $\mu$, i.e. $$\lim_n\int_I fg_ndm=\int_I f
d\mu
,\ \forall f\in C(I),$$
can we conclude that $\mu\perp \lambda$?
The result seems true to me and I was betting that the support of $\mu$ is contained in the set $$\{x\in I:\ \lim_n g_n(x)\neq 0\},$$
which I think, is equal to the sets $$\{x\in I:\ f(x)=\int_I fd\mu,\ \forall f\in C(I)\}=\{x\in I:\ \lim_n g_n(x)=\infty\}.$$
Any help is appreciated.
 A: EDIT: The following gives (the necessary ideas to construct) an example in which the weak-$\ast$-limit is (essentially) the Lebesgue on $[0,1]$, showing that $\mu \perp \lambda$ does in general not follow.
Try something along the following lines:
Fix $h \in C(\Bbb{R})$, $h\geq 0$, $\rm{supp}(h) \subset [-1,1]$ with $\int h \,dx = 1$.
For $i \in \Bbb{N}$ set $h^{(4^i)} := 4^i \cdot h(4^i \cdot)$.
For $f : \Bbb{R} \rightarrow \Bbb{C}$ let $(T_x f)(y) := f(y-x)$.
For $f \in C_c(\Bbb{R})$, we then have $\int (T_x h^{(4^i)})(y) \cdot f(y) \,dy \rightarrow f(x)$ for $i \rightarrow \infty$ uniformly in $x \in \Bbb{R}$ (using the uniform continuity of $f$).
[Here, one problem is that one would need this for $f \in C([0,1])$, when the integration is only over $[0,1]$ instead of $\Bbb{R}$.
EDIT: You can of course take something like $g_i := \sum_{j=1}^i \frac{1}{i} T_{\frac{1}{4} + \frac{j}{2i}}$ below. Then you will have $g_i \rightarrow \chi_{[1/4, 3/4]} \,dx$ and you dont have any problems, because the above convergence also holds for $f \in C([0,1])$ uniformly on $[1/4, 3/4]$.]
For $i \in \Bbb{N}$ let
$$
g_i := \sum_{j=1}^{i} \frac{1}{i} T_\frac{j}{i} h^{(4^i)}.
$$
Then $\int g_i \,dx= 1$ for all $i$.
Furthermore, (for $f \in C_c(\Bbb{R})$):
$$
\bigg| \int g_i(y) f(y) \,dy - \sum_{j=1}^{i} \frac{1}{i} f(\frac{j}{i}) \bigg| \leq \sum_{j=1}^{i}  \frac{1}{i} \bigg|\int (T_\frac{j}{i} h^{(4^i)})(y) f(y) \,dy - f(\frac{j}{i})\bigg|
\\ \leq \sup_x \bigg|\int (T_x h^{(4^i)})(y) f(y) \,dy - f(x) \bigg| \rightarrow 0
$$
for $i \rightarrow \infty$.
But note that $\sum_{j=1}^i \frac{1}{i} f(j/i) \rightarrow \int_0^1 f \,dx$.
This shows that $g_i \rightarrow dx$ (restricted to $[0,1]$) in the weak-$\ast$-sense.
But also note
$$
\lambda(\{x \mid g_i(x) \neq 0\}) \leq \sum_{j=1}^i \lambda(\{x \mid h^{(4^i)}(x) \neq 0\}) \leq 2 \cdot i \cdot 4^{-i},
$$
i.e. $\sum_i \lambda(\{x \mid g_i(x) \neq 0\}) < \infty$, where $\lambda$ denotes the Lebesgue measure. This easily implies $g_i \rightarrow 0$ almost everywhere.
