Below is a question from N.L. Carother's book Real Analysis. I've provided my attempt at a solutions, however, any feed back would be very appreciated.

Suppose $E$ is a measurable subset of $\mathbb{R}$ such that $m(E) = 1$. Show that:

(a) There is a measurable set $F$ with $F \subset E$ such that $m(F) = 1/2$.

(b) There is a closed set $F$, consisting entirely of irrationals, such that $F \subset E$ and $m(F) = 1/2$.

(c) There is a compact set $F$ with empty interior such that $F \subset E$ and $m(F) = 1/2$.

My Attempt:

(a) Define $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = m(V_x)$, where $V_x = E \cap (-\infty, x]$ for each $x $. It suffices to show that $f$ is an increasing continuous function and apply the Intermediate Value Theorem.

To show that $f$ is increasing, suppose that $x<y$ and note that $V_x \subset V_y$ so that, by the monotonicity of the Lebesgue measure, $f(x) = m(V_x) \leq m(V_y) = f(y)$.

Fix any $x \in \mathbb{R}$ and $\epsilon > 0$; choose $\delta = \epsilon$. Then, whenever $|x-y| < \delta$ with $x < y$, we have that

\begin{align} |f(x) - f(y)| \leq m(V_y \setminus V_x) \leq |x-y| < \delta = \epsilon, \end{align} which shows that $f$ is continuous.

Now, as $x$ increases, the sets $V_x$ increase to $E$. Hence $0 \leq f(x) \leq m(E) = 1$ for all $x$. Thus, by the Intermediate value theorem, there exists some $u \in \mathbb{R}$ for which $f(u) = 1/2$; that is, $m(V_u) = 1/2.$ Setting $F = V_u = E \cap (-\infty, u]\subset E$ is our desired measurable set.

(b) Let $(r_n)$ be an enumeration of the rationals. Define the set $R$ to be the union $$ R = \bigcup_{n=1}^{\infty} \left( r_n - \frac{1}{2^n}, r_n + \frac{1}{2^n} \right). $$ Once can see, without too much effort, that $m(R) = 2$; hence its complete, a closed set of infinite measure, $R^c$ as constructed consists of only irrationals.

*From here I've been pretty stuck; I'm still sitting on the ideas, however, I'm not sure where to go. Any HINTS would be of great asset. : )

  • 1
    $\begingroup$ To clarify, you want $m(E)$ to be $1$ and not $1/2$, right? $\endgroup$
    – Joey Zou
    Commented Jun 21, 2016 at 5:57
  • $\begingroup$ yes, i do. Thank you. $\endgroup$ Commented Jun 21, 2016 at 7:28
  • $\begingroup$ Are you allowed to used results on the regularity of the Lebesgue measure? Such results would be helpful.. $\endgroup$
    – Joey Zou
    Commented Jun 22, 2016 at 1:21
  • $\begingroup$ Your solution of the a) is beautiful! $\endgroup$ Commented Sep 9, 2021 at 3:21

1 Answer 1


I will show that there exists a compact $F\subset E$ consisting entirely of irrationals with $m(F) = \frac{1}{2}$. (The fact that this implies (b) and (c) should be an easy exercise.)

Since $E$ is measurable, $E\backslash\mathbb{Q}$ is as well, and $m(E\backslash\mathbb{Q}) = m(E) = 1$ since $m(\mathbb{Q}) = 0$. By inner regularity, there exists a compact $K\subset E\backslash\mathbb{Q}$ such that $m(K)>\frac{1}{2}$. Using a similar argument as presented in the original post, there exists $x>0$ such that $m(K\cap[-x,x]) = \frac{1}{2}$. Note that $K\cap[-x,x]$ is compact, and since $K\subset E\backslash\mathbb{Q}$ consists entirely of irrationals, $K\cap[-x,x]$ also consists entirely of irrationals. Thus, $F = K\cap[-x,x]$ is the desired compact set of irrationals with measure $\frac{1}{2}$.


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