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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.

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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.

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Thanks David; however, i have come across where they are saying that they are not mutually singular. They say that if we let $V_n$ be a decreasing sequence of open sets containing the rationals in $[0,1]$ with $m(V_n) < \frac1n$, then it's possible to find an appropriate $g_n$ with support in $V_n$ such that $\int fg_n dm$ is close to a Riemann sum for f, so that $\int fg_n dm\to\int f dm$. Any clarification? – Cecilia Apr 13 '12 at 12:14
In fact what I did seems to work if we assume that the $\{g_n\}$ are uniformly integrable, but it doesn't need to be the case (take $g_n=n\chi_{(0,1)}$ made continuous), but in this case we have convergence to $\delta_0$ and it works. I will think on the counter-example. – Davide Giraudo Apr 13 '12 at 20:30
Sure, i am also trying to come up with a counter example – Cecilia Apr 14 '12 at 7:38
Maybe an idea: take $g_n= n^2$ on $(k/n-\delta_n,k/n+\delta_n)$, piecewise linear on $(k/n-2\delta_n,k/n-\delta_n)$ and $(k/n+\delta_n,k/n+2\delta_n)$, and $0$ elsewhere. Then find $\delta_n$ in order to satisfy the conditions i) and ii). – Davide Giraudo Apr 14 '12 at 14:07
Isn't the uniformly integrable case vacuous? If $g_n \to 0$ a.e. and $g_n$ are uniformly integrable, then the Vitali convergence theorem would imply $\int g_n \to 0$, contradicting the assumption that $\int g_n = 1$. – Nate Eldredge Apr 14 '12 at 19:29

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