Exercise 6.9 in Rudin's RCA (Real and Complex Analysis)

Question

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?

Answer 1

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

1. 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 $k/n$, $0\leq k<n$.
2. 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 $\varepsilon>0$.
3. 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)
4. 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.

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

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Edit: As pointed out by @hdci, the sequence $g_n$ in fact converges to $0$ $m$-a.s. already. Thus, there is no need to extract a subsequence $g_{n'}$ with that property.

Here is prove that $g_n$ converges to $0$ $m$-a.s: Define $$S^*=\limsup_nS_n=\bigcap_{n\geq1}\bigcup_{k\geq n}S_k$$ $S^*$ is the set of $x\in I=[0,1]$ that belong to infinitely many $S_n$'s. Thus, for $x\in I\setminus S^*$, there is $n_x\in\mathbb{N}$ such that $x\in I\setminus S_n$ for all $n>n_x$ and so, by definition $g_n(x)=0$ for all $n>n_x$.

$$m(S^*)=\lim_nm\Big(\bigcup_{k\geq n}S_k\Big)\leq \lim_{n\rightarrow\infty}\sum_{k\geq n}k2^{-k}=0$$ This shows that $g_n\xrightarrow{n\rightarrow\infty}0$ $m$-a.s. in $I$.

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