Analysis Problem of the Day 30

Today’s problem is Problem 6 on the UCLA Fall 2024 Analysis Qual:

Problem 6: Let E \subseteq \mathbb{R} be a measurable set such that E=E+\frac{1}{n} for all positive integers n. Show that \mu(E)=0 or \mu(E^c)=0, where \mu is the Lebesgue measure.


Solution: Since the problem requires us to prove a statement with an exception of a measure zero set, we are likely to use measure theoretic machinery. In particular, notice that the above property of E in fact implies that E=E+\mathbb{Q}, i.e. E is \mathbb{Q}-invariant. The main tool in this problem turns out to be the Lebesgue differentiation theorem, which says that almost every point of a locally integrable function is a Lebesgue point, i.e. the average value of the function around the point converges to its value at that point.

Now, consider the function \chi_E. Pick x to be a Lebesgue point of \chi_E, and note that \chi_E(x)=\alpha where \alpha \in \{0,1\}. Either way, for any other Lebesgue point y \in \mathbb{R} and family of open balls B(y,r) centered at y, one has \int_{B(y,r)} \chi_E(x)dx = \int_{B(y-q_n,r)} \chi_E(x) dx \to \alpha as r \to 0, where q_n \in \mathbb{Q} are chosen so that y-q_n \to x and B(y-q_n,r) shrinks nicely to x as r \to 0. By the Lebesgue differentiation theorem, we conclude that \chi_E = \alpha at every Lebesgue point, that is, a.e. If \alpha=1, this implies \mu(E^c)=0, and otherwise \mu(E)=0, and we are done.

Remark: Note that this also implies that the set of Lebesgue points of E is precisely \chi_E^{-1}(\alpha).

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