Let \(f\) be a continuous function on \([0,1]\). Consider the following statements.
- \(f\) is absolutely continuous.
- For any \(\varepsilon>0\) there exists \(\delta>0\) such that \(m^*(f(E))<\varepsilon\) for any set \(E\subseteq[0,1]\) with \(m^*(E)<\delta\).
- \(m^*(f(E))=0\) for any set \(E\subseteq[0,1]\) with \(m^*(E)=0\).
Prove that (1) implies (2) and that (2) implies (3). Determine whether (3) implies (1).
The source asks to show that the three statements are equivalent for every continuous \(f\). That is false: a continuous function satisfying (3) need not have bounded variation, and so need not be absolutely continuous. Under the additional hypothesis that \(f\) has bounded variation, the three statements are equivalent; this is the Banach–Zarecki theorem. The source writes \(m\) for the measure of the arbitrary sets \(E\) and \(f(E)\); outer measure is used here because such sets need not be measurable.
Solution
1. Absolute continuity implies the small-image condition.
Assume \(f\) is absolutely continuous. Then \(f'\in L^1([0,1])\) and \[ f(y)-f(x)=\int_x^y f'(t)\,dt \] for all \(0\le x<y\le1\).
Fix \(\varepsilon>0\). By absolute continuity of the Lebesgue integral, choose \(\eta>0\) such that \[ m(U)<\eta\quad\Longrightarrow\quad \int_U|f'|<\varepsilon \] for every measurable \(U\subseteq[0,1]\). Put \(\delta=\eta/2\).
Let \(E\subseteq[0,1]\) with \(m^*(E)<\delta\). Choose an open set \(U\supseteq E\) with \(m(U)<\eta\). Write the relative open set \(U\cap[0,1]\) as a countable disjoint union of intervals \(I_j\). For each \(j\), \[ \operatorname{diam} f(I_j) \le \int_{I_j}|f'(t)|\,dt, \] because every difference \(|f(y)-f(x)|\) with \(x,y\in I_j\) is bounded by that integral. Hence \[ \begin{aligned} m^*(f(E)) &\le \sum_j m^*(f(E\cap I_j))\\ &\le \sum_j \operatorname{diam} f(I_j)\\ &\le \sum_j\int_{I_j}|f'|\\ &=\int_U|f'|<\varepsilon. \end{aligned} \] Thus (1) implies (2).
2. The small-image condition implies Lusin’s \(N\)-property.
Assume (2), and let \(E\subseteq[0,1]\) satisfy \(m(E)=0\). Given \(\varepsilon>0\), let \(\delta\) be supplied by (2). Since \(m(E)=0<\delta\), \[ m^*(f(E))<\varepsilon. \] As \(\varepsilon>0\) is arbitrary, \[ m^*(f(E))=0. \] Hence (2) implies (3).
3. Condition (3) does not imply absolute continuity.
Define \[ f(0)=0, \qquad f(x)=x\sin(x^{-2})\quad(0<x\le1). \] This function is continuous on \([0,1]\).
First we prove that it satisfies (3). Let \(E\subseteq[0,1]\) have measure zero and fix \(a\in(0,1)\). On \([a,1]\), the function \(f\) is \(C^1\), hence Lipschitz. Therefore \[ m^*(f(E\cap[a,1]))=0. \] On the other hand, \(|f(x)|\le x\), so \[ f(E\cap[0,a])\subseteq[-a,a] \] and therefore \[ m^*(f(E\cap[0,a]))\le2a. \] Consequently \[ m^*(f(E))\le2a. \] Letting \(a\downarrow0\) gives \(m^*(f(E))=0\). Thus \(f\) has Lusin’s \(N\)-property.
Now set \[ t_k=\left(\frac\pi2+k\pi\right)^{-1/2},\qquad k\ge0. \] Then \(t_k\downarrow0\) and \[ f(t_k)=(-1)^k t_k. \] For the partition containing \(0,t_N,t_{N-1},\ldots,t_0\) in increasing order, the variation is at least \[ \sum_{k=0}^{N-1}|f(t_k)-f(t_{k+1})| =\sum_{k=0}^{N-1}(t_k+t_{k+1}). \] Since \(t_k\asymp k^{-1/2}\), this tends to \(\infty\). Hence \(f\) has unbounded variation on \([0,1]\).
Every absolutely continuous function on a compact interval has bounded variation, so \(f\) is not absolutely continuous. Thus (3) does not imply (1).
4. For continuous \(f\) of bounded variation on \([0,1]\), conditions (1), (2), and (3) are equivalent.
The Banach–Zarecki theorem states that a real-valued function on a compact interval is absolutely continuous if and only if it is continuous, has bounded variation, and has Lusin’s \(N\)-property. For \(f\) of bounded variation this gives (3) implies (1), and steps 1 and 2 give the other implications. The function of step 3 has unbounded variation.