Mutually Singular measures c.f. Rudin's Real and Complex Analysis (Third Edition 1987) Chapter 6 Q9
Suppose that $\{g_n\}$ is a sequence of positive continuous functions on $I=[0,1]$, $\mu$ is a positive Borel measure on $I$, $m$ is the standard Lebesgue measure, and that
(i) $\lim_{n\to\infty}g_n(x)=0$ a.e. $[m]$
(ii) $\int_Ig_ndm=1$ for all $n$,
(iii) $\lim_{n\to\infty}\int_Ifg_ndm=\int_Ifd\mu$ for every $f\in C(I)$.
Does it follow that the measures $\mu$ and $m$ are mutually singular?
I know that $\mu$ and $m$ are mutually singular if they are concentrated in different disjoint sets, but how do I connect that with the 3 properties above? I will appreciate if someone can help me with the proof or counter example.
 A: Let $\delta_n:=\frac{2n}{(2n+1)n(n-1)}$, and $g_n$ defined by
$$g_n(x)=\begin{cases}
n&\mbox{ on }\left(\frac kn-\frac{\delta_n}2,\frac kn+\frac{\delta_n}2\right), 1\leq k\leq n-1\\\
\mbox{ linear }&\mbox{ on }\left(\frac kn-\frac{\delta_n}2-\frac{\delta_n}{2n},\frac kn-\frac{\delta_n}2\right)\\\
\mbox{ linear }&\mbox{ on }\left(\frac kn+\frac{\delta_n}2,\frac kn+\frac{\delta_n}2+\frac{\delta_n}{2n}\right)\\\
0&\mbox{ elsewhere}.
\end{cases}$$ 
We have that the measure of the support of $g_n$ is $(n-1)(1+1/n)\delta_n$ which converges to $0$, and $\int_{[0,1]}g_ndm=1$, except miscomputation. We can write 
$$\left|\int_{[0,1]}g_nfdm-\frac 1n\sum_{k=1}^nf\left(\frac kn\right)\right|\leq 2\delta_n \lVert f\rVert_{\infty}+\operatorname{mod}(f,\delta_n)\frac{n^2}{\delta_n},$$
where $\operatorname{mod}(f,\delta):=\sup\{|f(x)-f(y)|,x,y\in I, |x-y|\leq \delta \}$.
So the three conditions are full-filled by $m$ and $m$ are of course not singular.
A: It is possible to give a counterexample without having to construct it explicitly. It suffices to prove this for real continuous functions, since the definition of integration of complex functions and the linearity of the limit allow the extension of the result to all members of $C[0,1]$. Let $I_{n,j}= [\frac jn, \frac {j+2^{-n}}{n}]$ and define continuous positive functions $\phi_{n,j}$ on $I_{n,j}$ such that $$\int_{I_{n,j}} \phi_{n,j} = \frac 1n$$ and put $$h_n= \sum_{j=0}^{n-1}\phi_{n,j}$$
Notice $m\{x:h_n(x)\ne 0\}=\sum m(I_{n,j})=n\frac{2^{-n}}{n}=2^{-n} \to 0 \ \text{as} \ n \to \infty$. So that $h_n$ are non-negative continuous functions that converge to $0$ in measure, and so there is a subsequence (to be also denote by $h_n$ for sake of simplicity) 
along which $h_n$ converges to $0$ almost everywhere on $[0,1]$.

For any real continuous hence uniformly continuous $f$ on the compact unit interval, and any $j=0,...,n-1$ we have that $|f(x)-f(y)|<\epsilon$ with $x,y \in [\frac jn, \frac {j+1}n]$ (having chosen a sufficiently large $n$, of course).

By the above
$$f(\frac jn) -\epsilon<f(x)<\epsilon+f(\frac jn)$$
$$f(\frac jn)\phi_{n,j} -\epsilon\phi_{n,j}-f(\frac jn)\le f(x)\phi_{n,j}-f(\frac jn)\le\epsilon\phi_{n,j}+f(\frac jn)\phi_{n,j}-f(\frac jn)$$
Hence 
$$\int_{\frac jn}^{\frac {j+1}n}f(\frac jn)\phi_{n,j} -\epsilon\phi_{n,j}-f(\frac jn)\le \int_{\frac jn}^{\frac {j+1}n}f(x)\phi_{n,j}-f(\frac jn)\le\int_{\frac jn}^{\frac {j+1}n}\epsilon\phi_{n,j}+f(\frac jn)\phi_{n,j}-f(\frac jn)$$
Which implies 
$$\frac 1n \left(f(\frac jn) -\epsilon-f(\frac jn)\right)\le \int_{\frac jn}^{\frac {j+1}n}f(x)\phi_{n,j}-f(\frac jn)\le\frac 1n \left(f(\frac jn) +\epsilon-f(\frac jn)\right)$$
Summing over $j$ and noting that $\int_{\frac jn}^{\frac {j+1}n}f(x)\phi_{n,j}=\int_0^1f(x)\phi_{n,j}$ the above gives
$$-\epsilon \le \sum \int_0^1f(x)\phi_{n,j}-f(\frac jn) dx \le \epsilon$$ thus 
$$\left | \int_0^1 f(x)\sum \phi_{n,j}-\frac 1n \sum f(\frac jn)\right|\le \epsilon$$
