Analysis Problem of the Day 48

Today’s problem appeared as Problem 2 on Stanford’s Spring 2014 Analysis Qual

Problem 2. a) Show that there exists a continuous function $f:[0,1] \to \mathbb{R}$ and a null Lebesgue measurable subset $A \subset [0,1]$ such that $f(A)$ is measurable and $\mu(f(A))>0.$

b) Show that this isn’t possible if $f$ is absolutely continuous.

Solution: Let’s first show b). Indeed, $f$ is by definition absolutely continuous if and only if for all $\epsilon>0,$ there exists $\delta>0$ such that for all $n,$ $\sum_{i=0}^n |x_i-x_{i+1}|< \delta \implies \sum_{i=0}^{n} |f(x_i)-f(x_{i+1})| < \epsilon.$ By the definition of the Lebesgue measure as the restriction of an outer measure, if $A$ is a measure zero set, for any $\delta>0$ there exists a cover $\mathcal{U}$ of $A$ by countably many open intervals of total length less than $\delta.$ Notice that since the definition of absolute continuity holds for all finite sums, by extension it holds for countable sums. Thus, $f(\mathcal{U})$ is a cover of $f(A)$ with total measure less than $\epsilon$ (since for any open interval $(a,b)$ of length less than $\delta,$ $f((a,b))$ is contained in an open interval of length less than $\epsilon$ ). It follows that $\mu(f(A)) \leq \epsilon$ for any $\epsilon>0,$ i.e. $\mu(f(A))=0.$

To find such an $f$ in a), we must therefore require that $f$ is continuous but not absolutely continuous. A standard example is the Cantor function $C:[0,1] \to [0,1],$ which satisfies $f(\mathcal{C})=[0,1],$ so $\mu(f(\mathcal{C}))=1.$

Remark: Note that we may even require $f$ to be a homeomorphism. Then, $g=C(x)+x:[0,1] \to [0,2]$ is strictly increasing and continuous, therefore a homeomorphism. On every subinterval which does not intersect $\mathcal{C}$ (i.e. on which $f$ is locally constant), it follows that $\mu((g(a),g(b)))=f(b)+b-(f(a)+a)= b-a.$ It follows that $g$ preserves the measure of $\mathcal{C}^c,$ which is 1. Thus, $\mu(g(\mathcal{C}))=1.$ However, if $f$ is differentiable and the derivative is Lebesgue integrable, then the function is necessarily absolutely continuous.

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