# Singular functions

A monotone function $f$ on $[a,b]$ is called singular if $f'=0$ almost everywhere.

Let $f$ be a nondecreasing function on $[a,b]$ such that given $\epsilon~,~\delta\gt 0$, $\exists$ a finite collection $\{[y_k,x_k]\}$ of nonoverlapping intervals such that $$\sum |x_k-y_k|\lt \delta~~~~~\text{and}~~~~\sum\left(f(x_k)-f(y_k)\right)\gt f(b)-f(a)-\epsilon.$$
I would like to show that $f$ is singular.

Attempt:

From a previous exercise, I showed that a monotone function $f$ can be written as the sum of an absolutely continuous function, $g$ and a singular function, $h$. Thus $$f = g + h ,~~~\text{where}~~g=\int_a^x f' .$$
My goal is to show that $g=0$ almost everywhere. Let $I=\bigcup (y_k,x_k).$ Then $$\int_I f' = \sum \int_{(x_k,y_k)}~f' = \sum\left(f(x_k)-f(y_k)\right)\lt \epsilon.$$
Now, I know that since $f'$ is integrable, there is an $\epsilon$ such that $$\int_{[a,b]\setminus I}~f'\lt \epsilon.$$ But \begin{align*} 0\leq \int_a^b f' & = \int_I f'+\int_{[a,b]\setminus I}~f'\\ & \lt \epsilon + \epsilon\\ & = 2\epsilon. \end{align*}
Since $\epsilon$ is arbitrary, we can let $\epsilon \rightarrow 0$ and thus $$\int_a^b f'=0 ~~\text{and }~~g = 0.$$

Is what I've done right? Is there another way of approaching the problem?
Thanks.

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What are $(\alpha_i,\beta_i)$? –  Jonas Meyer Nov 5 '11 at 21:36
$\int_{(x_k,y_k)}~f' = f(x_k)-f(y_k)$ is incorrect; is this a typo? In any case, it is true that $\int_I f'$ can be made arbitrarily small, but this is by choosing $\delta$ small enough and using integrability of $f'$; I do not see justification in your solution. –  Jonas Meyer Nov 5 '11 at 21:39
$\bigcup(\alpha_i,\beta_i)=[a,b]\setminus I$...so how do I justify that there is such a $\delta$? –  Nana Nov 5 '11 at 22:23
$[a,b]\setminus I$ will not be a union of open intervals. (So you could just leave it as $[a,b]\setminus I$ or give it a shorter name if you want.) As for the question in your comment, see math.stackexchange.com/questions/40384/…. –  Jonas Meyer Nov 5 '11 at 22:27
okay. Thanks for the link. –  Nana Nov 5 '11 at 22:38

Given $\varepsilon>0$, there is a $\delta>0$ such that $\int_I f'<\varepsilon$ whenever $m(I)<\delta$. Thus you can choose $I=\cup(x_k,y_k)$ such that $\int_I f'<\varepsilon$ and $\sum_k f(y_k)-f(x_k)>f(b)-f(a)-\varepsilon$.
\begin{align*} &f(b)-f(a)-(h(b)-h(a))\\ &\leq f(b)-f(a) - (\sum_k h(y_k)-h(x_k))\\ &<\sum_k(g(y_k)-g(x_k))+\varepsilon\\ &=\int_I f' +\varepsilon\\ &<2\varepsilon. \end{align*}
Since $\varepsilon$ was arbitrary, this implies that $f(b)-f(a)=h(b)-h(a)$, so $g(b)-g(a)=0$, which in turn implies that $g=0$ everywhere. (Note how monotonicity is used here several times.)
Assume that there are $N$ intervals arranged so that $x_1<y_1\leq x_2<y_2\leq x_3<y_3\leq\cdots \leq x_N<y_N$. Then $(h(y_1)-h(x_1))+(h(y_2)-h(x_2))+\cdots+(h(y_N)-h(x_N))$ $=h(y_N)+(h(y_{N-1})-h(x_N))+\cdots+(h(y_1)-h(x_2))-h(x_1)$ $\leq h(y_N)-h(x_1)\leq h(b)-h(a)$ because $h$ is increasing. Perhaps more clearly (but less precisely), $h(b)-h(a)$ gives the total amount that $h$ changes, whereas $\sum_k h(y_k)-h(x_k)$ only counts the change that $h$ does on some nonoverlapping subintervals of $[a,b]$. –  Jonas Meyer Nov 14 '11 at 18:33