# Nondecreasing, singular function fails a weaker condition than absolutely continuous

Suppose $f:[0,1]\to\mathbb{R}$ is a nondecreasing function, and $f'(x)=0$ almost everywhere on $[0,1]$.

For any $\epsilon>0$, prove that there exists finitely many pairwise disjoint intervals $[a_k,b_k]$, $k=1,\dots, n$ in $[0,1]$ with $$\sum_{k=1}^n(b_k-a_k)<\epsilon$$ and $$\sum_{k=1}^n(f(b_k)-f(a_k))>f(1)-f(0)-\epsilon.$$

My attempt: I think proving by contradiction may be the way to go here.

Suppose to the contrary there exists $\epsilon>0$ such that for any finitely many pairwise disjoint $[a_k,b_k]\subseteq[0,1]$ with $\sum_{k=1}^n(b_k-a_k)<\epsilon$, we have $$0\leq\sum_{k=1}^n(f(b_k)-f(a_k))\leq f(1)-f(0)-\epsilon.$$

This looks like a weaker condition than absolutely continuous.

However, I am not sure how to proceed here, other than noting that $f(1)-f(0)>0$, so that $f$ is not constant. Also, I can see that $f$ cannot be absolutely continuous, otherwise it will have a contradiction, since an absolutely continuous and singular function must be constant.

Thanks for any help!

Update:

I have another idea, let $E$ be the set where $f'\neq 0$. Then $|E|=0$.

My idea is to cover $E$ in the Vitali sense by intervals $[a_k,b_k]$ and use Vitali Covering Theorem to prove that $\sum(b_k-a_k)<\epsilon$, but $[a_k, b_k]$ covers nearly all of $E$, where all the changes occurs, so that $\sum(f(b_k)-f(a_k))>f(1)-f(0)-\epsilon$.

At the moment, I have no idea how to make the above idea rigorous though.

• The conditions don't rule out $f(1) = f(0)$, but in that case it is trivial, $\sum f(b_k) - f(a_k) = 0 > -\varepsilon$. – Daniel Fischer Nov 24 '16 at 15:37
• Thanks. I meant that from the proof by contradiction, we have that $f(1)-f(0)\geq\epsilon>0$. – yoyostein Nov 24 '16 at 15:55

Let's extend $f$ to $[0, +\infty)$ by setting $f(x) = f(1)$ for $x > 1$. Then consider the functions $g_n \colon [0,1] \to \mathbb{R}$ defined by

$$g_n(x) = \frac{f\bigl(x + \frac{1}{n}\bigr) - f(x)}{\frac{1}{n}}.$$

By definition of the derivative, we have $g_n(x) \to f'(x)$ at all points where $f$ is differentiable, so by the hypothesis we have $g_n(x) \to 0$ almost everywhere. By Egorov's theorem, for every $\varepsilon > 0$, there is a measurable set $B_\varepsilon \subset [0,1]$ with $\lvert B_\varepsilon\rvert > 1 - \varepsilon$ such that the convergence $g_n(x) \to 0$ is uniform on $B_\varepsilon$. By the inner regularity of the Lebesgue measure, we can assume that $B_\varepsilon$ is compact. Then choose an $N$ such that $g_N(x) < \varepsilon$ for all $x \in B_\varepsilon$.

Now let $c_1 = \min B_\varepsilon$ and $d_1 = c_1 + \frac{1}{N}$. If we have already picked $c_i, d_i$ for $1 \leqslant i < k$, and $B_\varepsilon \cap (d_{k-1},1] \neq \varnothing$, let $c_k = \min (B_\varepsilon \cap [d_{k-1},1]$ and $d_k = c_k + \frac{1}{N}$.

We thus have $m \leqslant N$ pairs of points $0 \leqslant c_1 < d_1 \leqslant c_2 < d_2 \leqslant \dotsc \leqslant c_m < d_m$ with

$$B_\varepsilon \subset \bigcup_{k = 1}^m [c_k,d_k],$$

and

$$f(d_k) - f(c_k) = \frac{1}{N}g_N(c_k) < \frac{\varepsilon}{N}$$

for $1 \leqslant k \leqslant m$, so

$$\sum_{k = 1}^m \bigl(f(d_k) - f(c_k)\bigr) < m\frac{\varepsilon}{N} \leqslant \varepsilon.$$

Now let

$$U = [0,1] \setminus \bigcup_{k = 1}^m [c_k,d_k].$$

By construction $U$ consists of finitely many open [or half-open, if $c_1 > 0$, resp. $d_m < 1$, but that doesn't change anything important] intervals $(a_k,b_k)$, $1 \leqslant k \leqslant n$ (if $\varepsilon > f(1) - f(0)$, we could have $n = 0$, but in that case the assertion is trivial anyway) with

$$\sum_{k = 1}^n (b_k - a_k) \leqslant \lvert [0,1] \setminus B_\varepsilon\rvert < \varepsilon,$$

and

$$\sum_{k = 1}^n \bigl(f(b_k) - f(a_k)\bigr) = f(1) - f(0) - \sum_{k = 1}^m \bigl(f(d_k) - f(c_k)\bigr) > f(1) - f(0) - \varepsilon.$$

• Thanks. Amazing. Just to ask, what is the motivation or underlying idea behind this answer? (i.e. how would one think of this solution) – yoyostein Nov 25 '16 at 2:04
• Ignoring the trivial cases ($f(1) - f(0) < \varepsilon$), the task is to split $[0,1]$ into two subsets, each a finite union of intervals, such that one of the subsets is small, but almost all variation of $f$ happens there, and the other subset is large, but $f$ changes very little there. Now it's clear that $f$ changes very little in small neighbourhoods of a point where $f'$ vanishes, so we want to base our large set on such points. But to make sure that we can do with finitely many intervals, we don't want to allow these intervals to become too short. We can achieve that if we find a set – Daniel Fischer Nov 25 '16 at 15:27
• on which the average change of $f$ is uniformly small. The average change of $f$ between two points is the difference quotient, so we're led to look at a sequence of difference quotients, and want a large set on which that sequence converges uniformly. That points to Egorov's theorem. Once that idea is on the table, what remains is to shuffle the pieces until the jigsaw puzzle is assembled. – Daniel Fischer Nov 25 '16 at 15:27
• Thanks. I am amazed at how well the jigsaw puzzle fits together. – yoyostein Nov 25 '16 at 15:39