The Null-Set Preservation Principle
Absolute continuity controls the combined changes of a function across many short intervals. That control has a useful consequence: although a function can distort lengths, an absolutely continuous function on a compact interval cannot turn a set of Lebesgue measure zero into a set of positive measure. The proof uses open interval covers and sums of increments; it does not require the null set itself to be a union of intervals.
Throughout, \(a\le b\) are real numbers, \(m\) is Lebesgue measure on \(\mathbb{R}\), and \(f:[a,b]\to\mathbb{R}\) is real-valued. A set \(N\subseteq[a,b]\) is called Lebesgue-null if its Lebesgue outer measure \(m^*(N)\) is zero. In particular, \(N\) is Lebesgue measurable because Lebesgue measure is complete. We use absolute continuity of a function, not absolute continuity of one measure with respect to another.
The target statement concerns \(f(N)=\{f(x):x\in N\}\). It is not necessary to assume that \(f\) is one-to-one or monotone. We will cover \(f(N)\) by intervals whose total lengths can be made arbitrarily small.
Controlling Oscillation on Open Components
For a nonempty set \(C\subseteq[a,b]\), define its oscillation under \(f\) by \(\operatorname{osc}(f,C)=\sup\{|f(x)-f(y)|:x,y\in C\}\). This is the length of the smallest closed interval containing \(f(C)\), when the oscillation is finite. The key point is that absolute continuity controls the sum of these oscillations over the separate components of a small open set.
Proof. First, each component \(C_j\) has finite oscillation. For any \(x,y\in C_j\), the interval with endpoints \(x\) and \(y\) lies in \(C_j\), and its length is at most \(m(U)<\delta\). The absolute-continuity condition applied to this one interval gives \(|f(x)-f(y)|<\varepsilon\). Consequently \(\operatorname{osc}(f,C_j)\le\varepsilon\).
Now take any finite collection of distinct components, say \(C_{j_1},\ldots,C_{j_r}\). For each \(i\), choose \(x_i,y_i\in C_{j_i}\) so that \(|f(x_i)-f(y_i)|\) is as close as desired from below to \(\operatorname{osc}(f,C_{j_i})\). If an oscillation is zero, it contributes zero and can be omitted. The intervals with endpoints \(x_i,y_i\) lie in distinct components, so they are pairwise disjoint. Their total length is at most \(m(U)<\delta\). Absolute continuity therefore gives $$ \sum_{i=1}^r |f(x_i)-f(y_i)|<\varepsilon. $$ Approximating each of the finitely many oscillations arbitrarily closely shows that \(\sum_{i=1}^r\operatorname{osc}(f,C_{j_i})\le\varepsilon\). This holds for every finite collection. Since a relatively open subset of an interval has at most countably many components, the sum over all components, defined as the supremum of its finite partial sums, is at most \(\varepsilon\). \(\square\)
Countability here follows because each nonempty component with positive length contains a rational number, and distinct components contain distinct rationals; any components meeting an endpoint also contain points of positive distance from that endpoint. The lemma does not assert that the image of a component is itself an interval. Instead, it uses the closed interval from \(\inf f(C_j)\) to \(\sup f(C_j)\), whose length is exactly the oscillation.
Absolutely Continuous Maps Preserve Null Sets
Proof. If \(a=b\), then \([a,b]\) has only one point and \(f(N)\) is empty or a singleton, so its outer measure is zero. Assume \(a<b\), and fix an arbitrary \(\eta>0\). Use absolute continuity with \(\varepsilon=\eta/2\) to obtain \(\delta>0\). Since \(m^*(N)=0\), the characterization of Lebesgue-null sets by open covers gives an open set \(O\subseteq\mathbb{R}\) containing \(N\) such that \(m(O)<\delta\). Set \(U=O\cap[a,b]\). Then \(U\) is relatively open in \([a,b]\), contains \(N\), and satisfies \(m(U)<\delta\).
Let \(C_j\) be the nonempty components of \(U\). Each \(x\in N\) lies in one of these components, so \(f(N)\subseteq\bigcup_j f(C_j)\). For each \(j\), the lemma gives a finite oscillation, and \(f(C_j)\) is contained in the closed interval \([\inf f(C_j),\sup f(C_j)]\), of length \(\operatorname{osc}(f,C_j)\). Thus these intervals cover \(f(N)\). Countable subadditivity of Lebesgue outer measure and the lemma imply $$ m^*(f(N)) \le \sum_j\operatorname{osc}(f,C_j) \le \frac{\eta}{2} <\eta. $$ The number \(\eta>0\) was arbitrary, so \(m^*(f(N))=0\). Completeness of Lebesgue measure then implies that \(f(N)\) is Lebesgue measurable and has measure zero. \(\square\)
Worked Applications
Worked Example: A Linear Map
Let \(f:[0,1]\to\mathbb{R}\) be \(f(x)=3x-2\), and let \(N\subseteq[0,1]\) be Lebesgue-null. First check absolute continuity directly. For any finite pairwise disjoint intervals \([u_j,v_j]\), $$ \sum_j|f(v_j)-f(u_j)| =\sum_j|3v_j-2-(3u_j-2)| =3\sum_j(v_j-u_j). $$ Given \(\varepsilon>0\), choosing \(\delta=\varepsilon/3\) makes the increment sum less than \(\varepsilon\) whenever the total interval length is less than \(\delta\).
For a concrete cover estimate, take intervals \(I_j\) covering \(N\) with \(\sum_j |I_j|<1/100\); such a cover exists because \(m^*(N)=0\). The image of each interval under \(x\mapsto3x-2\) is an interval of length \(3|I_j|\). Hence these image intervals cover \(f(N)\), and $$ m^*(f(N))\le\sum_j3|I_j|=3\sum_j|I_j|<\frac{3}{100}. $$ More generally, using covers with total length less than any prescribed positive number proves \(m^*(f(N))=0\). Check: the scale factor is \(3\), since \(f(v)-f(u)=3(v-u)\); the displayed bound is \(3(1/100)=3/100\).
Worked Example: Squaring a Countable Null Set
Let \(f:[0,1]\to\mathbb{R}\) be \(f(x)=x^2\), and take \(N=\mathbb{Q}\cap[0,1]\), a countable set and therefore Lebesgue-null. For \(0\le u\le v\le1\), $$ |v^2-u^2|=(v-u)(v+u)\le2(v-u), $$ because \(v+u\le2\). Summing this inequality over any finite pairwise disjoint family of intervals shows that \(f\) is absolutely continuous: given \(\varepsilon>0\), take \(\delta=\varepsilon/2\).
There is also a direct cover calculation. Enumerate the elements of \(N\) as \(q_1,q_2,\ldots\). For any \(\eta>0\), cover \(q_n^2\) by an open interval of radius \(\eta/2^{n+2}\). The length of the \(n\)th interval is \(2(\eta/2^{n+2})=\eta/2^{n+1}\), so the total length is $$ \sum_{n=1}^{\infty}\frac{\eta}{2^{n+1}} =\eta\left(\frac14+\frac18+\frac1{16}+\cdots\right) =\frac{\eta}{2}. $$ These intervals cover \(f(N)\), so \(m^*(f(N))\le\eta/2\) for every \(\eta>0\), and therefore \(m^*(f(N))=0\). Check: the geometric series in parentheses is \((1/4)/(1-1/2)=1/2\), giving the stated total \(\eta/2\).
Worked Example: A Tent Function and a Misleading Endpoint Increment
Define \(T:[0,1]\to\mathbb{R}\) by \(T(x)=2x\) for \(0\le x\le1/2\) and \(T(x)=2-2x\) for \(1/2\le x\le1\). On either linear piece, \(|T(v)-T(u)|=2|v-u|\). If \(u\le1/2\le v\), the triangle inequality gives $$ |T(v)-T(u)| \le |T(1/2)-T(u)|+|T(v)-T(1/2)| =2(1/2-u)+2(v-1/2) =2(v-u). $$ Thus \(T\) is Lipschitz with constant \(2\), and summing this bound over disjoint intervals proves absolute continuity with \(\delta=\varepsilon/2\).
For \(N=\{0,1/2,1\}\), the image is \(T(N)=\{0,1\}\), which is null. For any \(r>0\), intervals of radius \(r\) about \(0\) and \(1\) cover that image and have total length \(4r\); for example, \(r=1/100\) gives total length \(4/100=1/25\). Check: the image points are \(T(0)=0\), \(T(1/2)=1\), and \(T(1)=0\).
This example also exposes a common error. One might incorrectly claim that the length of \(T([0,1])\) is \(|T(1)-T(0)|=0\). In fact, \(T([0,1])=[0,1]\), of length \(1\). The endpoint increment misses the rise and fall inside the interval. The correction is to control oscillation using increments between points in the interval: here \(T(1/2)-T(0)=1\) and \(T(1)-T(1/2)=-1\), so the sum of the absolute increments is \(1+1=2\), not zero.
What the Result Does—and Does Not—Say
The theorem is a statement about images of null sets, not a claim that lengths are preserved. A function may stretch intervals by a large factor and still preserve null sets, provided its total increments satisfy the absolute-continuity condition. The proof works because a null set can be placed inside an open set of arbitrarily small measure, and absolute continuity bounds the combined oscillation over that open set's components.
It is important not to replace absolute continuity by continuity in the argument. Continuity controls the change over one sufficiently short interval, but it does not provide the required bound on the sum of changes over a finite family of disjoint intervals. Nor does the theorem say that \(f(N)\) has to be an interval: it says that its Lebesgue outer measure is zero. Completeness of Lebesgue measure then supplies measurability.
Check Your Understanding
Use the interval-increment definition and the covering argument to justify each answer.
- For \(f(x)=5x+1\) on \([0,1]\), what choice of \(\delta\) verifies absolute continuity for a given \(\varepsilon>0\)?
- Why can each \(f(C_j)\) be covered by an interval of length \(\operatorname{osc}(f,C_j)\), even if \(f(C_j)\) is not an interval?
- Where in the proof of null-set preservation is completeness of Lebesgue measure used?
- For the tent function \(T\) above, calculate \(T(1)-T(0)\) and compare it with the oscillation of \(T\) on \([0,1]\).
- Explain why the bound on the sum of component oscillations is obtained first for finite collections before being extended to countably many components.