# Intermediate Value Like Property for Lebesgue Measure

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 $$x and note that $$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. : )

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

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 $\frac{1}{2}$.