[Math] Mutually Singular measures

Question

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.

Answer 1

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.