Fall 2009 - Problem 5

Solution.

Let $\set{r_n}_n$ be an enumeration of $\Q$. Define $I_n = \p{r_n - 2^{-n}, r_n + 2^{-n}}$,

$G_n$ is open since it is a union of open sets. Also $G_n$ contains all but possibly finitely many rational numbers, so it is dense. By construction, $G_{n+1} \subseteq G_n$ and

Fix $k \geq 1$. We will construct disjoint sets which remove a small, positive measure portion of $I_k$.

By density of $\Q$ again, for any $n \geq 1$, there exists $j \geq n$ such that $q_j \in I_k \implies I_j \cap I_k \neq \emptyset$. Since $I_j \subseteq G_n$,

Thus, since $m\p{G_\ell} \to 0$ as $\ell \to \infty$, there exists $\ell > n$ such that

since $G_\ell \subseteq G_n$. Intuitively, any interval $I_k$ will intersect with $G_n$ with positive measure. To get infinitely many of these, we can remove the tail $G_\ell$ of $G_n$ and continue inductively.

Let $n_1 = 1$, and inductively pick $n_1 < n_2 < \cdots$ as follows: Given $n_{2k-1}$,

Moreover, because $n \subseteq m \implies G_m \subseteq G_n$, it follows that the $G_{n_{2k-1}} \setminus G_{n_{2k}}$ are pairwise disjoint. Set

Now let $I = \br{a, b} \subseteq \R$ be such that $a < b$. If $n$ is large enough, then $\br{a + 2^{-n}, b - 2^{-n}} \neq \emptyset$. Thus, by density, there exists $k \geq n$ such that $q_k \in \br{a + 2^{-n}, b - 2^{-n}}$, and so

which means $I_k \subseteq I$.

Notice that $E \cap \p{G_{n_{2k}} \setminus G_{n_{2k+1}}} = \emptyset$ by construction. Indeed, $G_{n_{2\ell-1}}$ is always odd, whereas $G_{n_{2k}}$ is even, and the $G_{n_{m}} \setminus G_{n_{m+1}}$ are pairwise disjoint. Consequently, $G_{n_{2k}} \setminus G_{n_{2k+1}} \subseteq E^\comp$, which means

On the other hand,

Hence, $0 < m\p{E \cap I} < m\p{I}$, which was what we wanted.