Exercise 6.9 in Rudin's RCA (Real and Complex Analysis) The following is an exercise 6.9 in Rudin's Real and Complex Analysis:
Suppose that $\{ g_n \}$ is a sequence of positive continuous functions on $I=[0,1]$, that $\mu$ is a positive Borel measure on $I$, and that 
(i) lim$_{n\to \infty}$ $g_n (x) = 0$ a.e. [m], 
(ii) $\int_I g_n dm = 1$ for all $n$, 
(iii) lim$_{n\to \infty}$ $\int_I fg_n dm = \int_I f d\mu$ for every $f\in C(I)$.
Does it follow that $\mu \perp m$?


*

*I think that the answer is positive. $\{g_n\}$ seems to be something similar to good kernel. I tried to use Egoroff's theorem and then derive something useful, but couldn't. Would you please give me some help?

 A: This counter example that shows that the answer is no in general:

*

*For each integer $n$ define $g_n$ to be a nonnegative continuous function on $[0,1]$ which equals $0$ except on intervals of the form $I_{n,k}=\big[\frac{k}{n},\frac{k}{n}+\frac{1}{2^n}\big]$, $0\leq k<n$, where
$$\int_{I_{nk}}g_n=\frac1n$$
One can for instance use piecewise linear functions that take large values at points $k/n$, $0\leq k<n$.

*This sequence converges to $0$ in measure: $m\big(|g_n|>\varepsilon\big)\leq m\big(g_n\neq0\big)\leq n 2^{-n}\xrightarrow{n\rightarrow\infty}0$ for all $\varepsilon>0$.

*One can then take a subsequent $g_{n'}$ which converges to $0$ $m$-a.s. For sake of simplicity, let us denote the subsequence as $g_n$ (abuse of notation if you will)

*Claim: Let $f\in\mathcal{C}[0,1]$.
$$\lim_n\int f g_n\,dm = \int f\,dm\tag{1}\label{one}$$
One can appeal to uniform continuity to get for any $\varepsilon >0$, a $\delta>0$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. Taking $n$ sufficiently large (e.g. $n>\frac{1}{\delta}$)

$$
\Big(f\big(\frac{k}{n}\big) -\varepsilon\Big)g_n\mathbb{1}_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]}\leq f g_n\mathbb{1}_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]}\leq \Big(\varepsilon  + f\big(\frac{k}{n}\big)\Big)g_n\mathbb{1}_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]}
$$
Integration gives
$$
\Big( f\big(\frac{k}{n}\big) -\varepsilon\Big)\frac{1}{n}\leq \int_{\big[\tfrac{k}{n},\tfrac{k+1}{n}\big]} f g_n\,dm \leq \Big(\varepsilon  + f\big(\frac{k}{n}\big)\Big)\frac{1}{n}
$$
Adding over all $0\leq k<n$ results in
$$
\Big|\int_I fg_n\,dm -\sum^{n-1}_{k=0}f\big(\frac{k}{n}\big)\frac{1}{n}\Big|\leq\varepsilon
$$
As $f$ is continuous, $\sum^{n-1}_{k=0}f\big(\frac{k}{n}\big)\frac{1}{n}\xrightarrow{n\rightarrow\infty}\int_If\,dm$ and the claim follows.


*This gives a negative answer to the question in Rudin's problem by taking $\mu=m$.

