Nondecreasing, singular function fails a weaker condition than absolutely continuous

Question

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 $$\sum_{k=1}^n(f(b_k)-f(a_k))>f(1)-f(0)-\epsilon.$$

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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!

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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.

Answer 1

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

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

Answer 2

Instead of using Egorov's Theorem, I adopt the strategy of using the Vitali Covering lemma. Let's start our argument below.

Define the set \begin{align} \Lambda:=\{f'=0\} \end{align} Then for all $x\in\Lambda$, there exists $h^x_k>0$ with $h^x_k\rightarrow 0$ such that \begin{align} \frac{f(x+h^x_k)-f(x)}{h^x_k}<\epsilon \end{align} Consider that \begin{align}\mathscr{B}:=\{[x,x+h^x_k]:x\in\Lambda,k\in\mathbb{N}\}, \end{align} which is a Vitali Covering of $[0,1]$. Then there are disjoint intervals $I^x_k$ which is finitely many, says m, such that \begin{align} (1). \sum|I^x_k|\leq 1+\delta \end{align} and \begin{align} (2). m^{*}([0,1]-\bigsqcup I^x_k)<\delta\ \end{align} The idea of using the Vitali Covering lemma is that it can select the representative intervals where the difference of the function $f$ can be controlled. Besides, the measure of other intervals is quite small and the function $f$ on it represents almost all of its value. This is exactly what we want to prove.
Write $[0,1]-\bigsqcup I^x_k=\bigsqcup (x_k,y_k)$, and let the number of intervals $(x_k,y_k)$ to be $N$. So, we have \begin{align}f(1)-f(0)=\sum^{N}_{k=1}(f(y_k)-f(x_k))+\sum Diff_f(I^x_k)\leq\sum^{N}_{k=1}(f(y_k)-f(x_k))+\epsilon,\end{align} which implies that \begin{align}f(b)-f(a)-\epsilon\leq\sum^{N}_{k=1}(f(y_k)-f(x_k))\end{align}

The symbol $Diff_f(I^x_k)$ means the value of subtracting the value of $f$ on the endpoints of $I^x_k$.