June 14, 2026
Problem 2.14 in Probability and Measure by Billingsley requires knowing the definition of nowhere-dense sets. Appendix 15 provides a definition of nowhere-dense sets and sets of the first category with respect to \(\mathbb{R}\):
The set \(A\) is, by definition, dense in the set \(B\) if for each \(x \in B\) and each open interval \(J\) containing \(x\), \(J\) meets \(A\). This is the same thing as requiring \(B \subseteq \overline{A}\) (the closure of \(A\)). The set \(E\) is, by definition, nowhere dense if each open interval \(I\) contains some open interval \(J\) that does not meet \(E\). This makes sense: if \(I\) contains an open interval \(J\) that does not meet \(E\), then \(E\) is not dense in \(I\); the definition requires that \(E\) be dense in no interval \(I\).
The usual topological definition of density is as follows: Let \((X, d_X)\) be a metric space. Let \(A, B \subset X\). We say that \(A\) is dense in \(B\) if and only if \(\overline{A} = B\). The topological definition of nowhere-dense is: \(A\) is nowhere-dense in \(X\) if \((\overline{A})^{\mathrm{o}} = \emptyset\). We show that these definitions are consistent with the ones given in the textbook when \(X = \mathbb{R}\).
Let's show that the definition of density is consistent. First, we suppose that \(A\) is dense in \(B\) if and only if for every \(x \in B\) and every open interval \(J\) containing \(x\), \(J \cap A \ne \emptyset\). We show that this means \(\overline{A} = B\). Let's show that \(B \subset \overline{A}\). Let \(x \in B\). We show that \(x \in \overline{A}\). There exists open interval \(J\), \(x \in J\), such that \(J \cap A \ne \emptyset\). If \(x \in J \cap A\), then \(x \in \overline{A}\) (and so we're done). Otherwise, if \(x \not\in J \cap A\), then \((J \setminus \{x\}) \cap A \ne \emptyset\). Since this is true for every open interval \(J\) containing \(x\), \(x \in A'\) (where \(A'\) denotes the limit points of \(A\)). Therefore, \(x \in \overline{A}\). Now we show that \(\overline{A} \subset B\). Let \(x \in \overline{A}\). For every open interval \(x \in J\), \(J \cap A \ne \emptyset\). This means \(x \in B\) ; if not, then \(A\) would not be dense in \(B\) (because for every \(x \in B\) and every open interval \(J\) containing \(x\), \(J \cap A \ne \emptyset\)). So, \(\overline{A} = B\).
Now, let's suppose that \(\overline{A} = B\). Then, we show that for every \(x \in B\) and every open interval \(J\) containing \(x\), \(J \cap A \ne \emptyset\). Let \(x \in B\). Then, \(x \in \overline{A}\). So, \(x \in A\) or \(x \in A'\). Choose an arbitrary open interval \(J\) containing \(x\). If \(x \in A\), then \(J \cap A \ne \emptyset\) because \(x \in J \cap A\). Otherwise, if \(x \in A'\), then \((J \setminus \{x\}) \cap A \ne \emptyset\). This immediately implies \(J \cap A \ne \emptyset\).
This proves that the definitions of density are equivalent (for \(X = \mathbb{R}\)). Now, we show that the definition of nowhere-dense is consistent when \(X = \mathbb{R}\). First, let's clarify the definition given by Billingsley: \(A\) is nowhere-dense if for every open interval \(I\), there exists open interval \(J \subset I\) so that \(J \cap A = \emptyset\). This is the same thing as saying \(A\) is not dense in any open interval \(I\). To see this, note that if \(A\) is not dense in any \(I\) (nowhere-dense), then for some \(x \in I\) and some open interval \(J\) containing \(x\), \(J \cap A = \emptyset\). Then, for any open interval \(J' \subset J\), note that \(J' \cap A = \emptyset\). So, if we take this interval small enough, we can find a \(J' \subset I\) so that \(J' \cap A = \emptyset\).
Now, we show that the topological definition is consistent. First, suppose that Billingsley's definition is true. We show that \((\overline{A})^{\mathrm{o}} = \emptyset\). Suppose for contradiction that \(x \in \overline{A}^{\mathrm{o}}\). Then, there exists open interval \(I\) such that \(x \in I\) and \(I \subset \overline{A}\). Since \(A\) is nowhere-dense, there exists open interval \(J \subset I\) such that \(J \cap A = \emptyset\). But \(J \subset \overline{A}\). Since \(J \cap A = \emptyset\), that means \(J \subset A'\). Consider any \(y \in J\). Then, \(y \in A'\). Since \(J\) is open, \((J \setminus \{y\}) \cap A \ne \emptyset\). This implies that \(J \cap A \ne \emptyset\), which is a contradiction. So, \((\overline{A})^{\mathrm{o}} = \emptyset\).
Next, let's prove the other direction: suppose that \((\overline{A})^{\mathrm{o}} = \emptyset\). Suppose there exists open interval \(I\) such that for every open interval \(J\), \(J \subset I\), \(J \cap A \ne \emptyset\). Choose any \(x \in I\). For every open \(J\) containing \(x\), \(J \cap A \ne \emptyset\). Then, either \(x \in A\) or \(x \in A'\). This implies that \(x \in \overline{A}\). Since \(x \in I\) was arbitrary, this means that \(I \subset \overline{A}\). For any \(x\), \(I\) is an open neighborhood contained in \(\overline{A}\), so \((\overline{A})^{\mathrm{o}} \ne \emptyset\). But this contradicts the hypothesis, so we must have that for every open interval \(I\), there exists an open interval \(J\), \(J \subset I\), such that \(J \cap A = \emptyset\).
This shows that the topological definition of nowhere-dense is the same as the definition given by Billingsley when we take \(X = \mathbb{R}\).