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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.

J.R.
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Cecilia
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2 Answers2

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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.

Davide Giraudo
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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 07:38
  • 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
  • But Davide,${g_n}$ are not necessarily uniform intergrable just as you also noted earlier on. – Cecilia Apr 14 '12 at 17:15
  • Yes, and that's why we can provide a counter-example. – Davide Giraudo Apr 14 '12 at 17:43
  • Sure. I am waiting for the counter example. – Cecilia Apr 14 '12 at 18:12
  • I've edited my initial post. – Davide Giraudo Apr 14 '12 at 19:07
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    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
  • @NateEldredge Indeed, I didn't think to use this theorem. So I can deleted a great part of my "answer". Thanks! – Davide Giraudo Apr 14 '12 at 20:06
  • I think $\delta_n$ should be $2/(n-1)(2n+1)$. Since $\int g_n=(\delta_n+\delta_n+\delta_n/n)(n/2)(n-1)$ – mnmn1993 Dec 22 '17 at 13:59
  • Also,may I have a proof of your inequality? I can only get this form For each interval $I_k=(\dfrac{k}{n}-\dfrac{\delta_n}{2}-\dfrac{\delta_n}{2n},\dfrac{k}{n}+\dfrac{\delta_n}{2}+\dfrac{\delta_n}{2n})$, $\int_{I_k} g_n=1/(n-1)$ \begin{align} \left|\int_{I_k}fg_n-\dfrac{1}{n}f(k/n)\right|&= \left|\int_{I_k}g_n(f-\dfrac{n-1}{n}f(\dfrac{k}{n}))\right| \ &\leq \left|\int_{I_k}g_n(f-f(\dfrac{k}{n}))\right|+\left|\int_{I_k}g_n\dfrac{1}{n}f(\dfrac{k}{n})\right|\ &=\dfrac{1}{n-1}mod(f,\delta_n)+|f|_{\infty}\dfrac{1}{n(n-1)} \end{align} – mnmn1993 Dec 22 '17 at 14:38
  • $$\left|\int_0^1fg_n-\dfrac{1}{n}\sum^n_1f(k/n)\right|\leq \sum^{n-1}1\left|\int{I_k}g_n(f-\dfrac{n-1}{n}f(\dfrac{k}{n}))\right|+f(1)/n\leq mod(f,\delta_n)+|f|_{\infty}\dfrac{1}{n}+f(1)/n$$ – mnmn1993 Dec 22 '17 at 14:41
  • @ Davide Giraudo: It is not clear that $g_n$ converges to $0$ a.s. in fact, $0=\liminf_ng_n<\liminf_ng_n=1$ a.s. The sequence does converge to $0$ in measure however, after which we can pass to a subsequence to get a.s. convergence. – Mittens Jun 05 '20 at 22:38
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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 $$\left | \int_0^1 f(x)\sum \phi_{n,j}-\frac 1n \sum f(\frac jn)\right|\le \epsilon$$

Mittens
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GuPe
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  • The sequence $h_n$ does not converge to $0$ a.s. necessarily. It does however converge to $0$ in measure: $m\Big(|h_n|>\varepsilon\Big)\leq m\Big(h_n\neq0\Big)\xrightarrow{n\rightarrow\infty}0$. Then one can pass to a subsequence $h_{n'}$ that converges to $0$ a..=s. – Mittens Jun 05 '20 at 21:00