$M \subseteq \mathbb R$ exist, so that $\lambda^{1}(M \cap I)= \frac{1}{2}\lambda^{1}(I)$ for any bounded Interval $I\subseteq \mathbb R$.

Question

Does a Lebesgue-measurable Subset $M \subseteq \mathbb R$ exist, so that $\lambda^{1}(M \cap I)= \frac{1}{2}\lambda^{1}(I)$ for any bounded Interval $I\subseteq \mathbb R$.

My idea:

Assume that such a lebesgue-measurable set $M$ did exist, it then follows that for any $\epsilon > 0$ we can find $O_{\epsilon}$ open and $C_{\epsilon}$ closed so that $\lambda^{1}(O_{\epsilon} \setminus C_{\epsilon})< \epsilon$ and $C_{\epsilon} \subseteq M \subseteq O_{\epsilon}$

it then follows for arbitrary bounded sets $I$ that $\lambda^{1}(C_{\epsilon}\cap I)\leq \frac{1}{2} \lambda^{1}(I)\leq \lambda^{1}(O_{\epsilon}\cap I)$

I am struggling to find any contradictions that I can use. Any ideas?

Answer 1

Approach 1: Use Lebesgue's differentiation theorem.

$$\frac{1}{\lambda(I)} \lambda(I \cap M) = \frac{1}{\lambda(I)} \int_I 1_M(x) \, \lambda(dx).$$

Now take $I=(x_0-r,x_0+r)$ for fixed $x_0 \in \mathbb{R}$. What happens if we let $r \to 0$?

Approach 2: Use outer regularity of Lebesgue measure.

Suppose there were such a set $M$. It is immediate that the set has strictly positive Lebesgue measure (otherwise $\lambda(I \cap M)$ would be zero for all $I$); therefore we may without loss of generality that $M' := M \cap [0,1]$ has positive Lebesgue measure. For fixed $\epsilon>0$ there exists an open set $O \supseteq M'$ such that $\lambda(O \setminus M') \leq \epsilon$. Write $O=\bigcup_{k \in \mathbb{N}} I_k$ as union of bounded intervals; we can choose the intervals in such a way that they do not overlap too much, in the sense that, say, $\sum_{k \geq 1} \lambda(I_k) \leq \frac{3}{2} \lambda(O)$. Then \begin{align*} \lambda(M') = \lambda(O \cap M') &\leq \sum_{k \geq 1} \lambda(I_k \cap M') \\ &\leq \sum_{k \geq 1} \lambda(I_k \cap M) \\ &= \frac{1}{2} \sum_{k \geq 1} \lambda(I_k) \\ &\leq \frac{3}{4} \lambda(O) \\ &\leq \frac{3}{4} (\lambda(M')+\epsilon).\end{align*}

If we choose $\epsilon$ sufficiently small, e.g. $\epsilon=\frac{1}{4} \lambda(M')>0$, this gives a contradiction:$$0<\lambda(M')< \frac{15}{16} \lambda(M').$$