A non-measurable set with density $\frac{1}{2}$

lebesgue-measurereal-analysis

According to Lebesgue's Density Theorem:

Let $\mu$ be the Lebesgue outer measure, and let $A\subseteq\mathbb{R}$ be a Lebesgue measurable set. Then the limit: $$\lim_{h\to0}\frac{\mu\left(A\cap\left(x-h,x+h\right)\right)}{2h}$$ exists and equals to either 0 or 1, for almost every $x\in\mathbb{R}$.

Is there a known example of a set $A\subseteq\mathbb{R}$ for which the claim above doesn't hold?

If not, Is there a known proof of existance of such a set?

(A set like this, if exists, is clearly non-measurable by the theorem above).

Best Answer

The answer is no.

Following the notation in the question, I will use $\mu$ to denote the Lebesgue outer measure; I will use $|\cdot|$ to denote the Lebesgue measure.

We will use the following standard result:

Standard Result Given any set $A$, there exists a $G_\delta$ set $\hat{A} \supseteq A$ such that for every Lebesgue measurable $M$, we have $\mu(A\cap M) = |\hat{A}\cap M|$.

Sketch of proof: for each $n$, let $A_n$ denote an open superset of $A$ with $\mu(A_n) \leq \mu(A) + \frac1n$. Let $\hat{A} = \cap_n A_n$. This shows that $\hat{A}$ is measurable and $\mu(\hat{A}) = \mu(A)$. Given $M$ measurable we note that $\mu(A) = \mu(A\cap M) + \mu(A\cap M^c)$. Combining $\mu(\hat{A}\cap M) + \mu(\hat{A}\cap M^c) = \mu(A\cap M) + \mu(A\cap M^c)$ and the inequalities $\mu(A\cap M) \leq \mu(\hat{A}\cap M)$ and $\mu(A\cap M^c) \leq \mu(\hat{A}\cap M^c)$ we get the result. QED

Given a set $A$ and a point $x$, denote by $$ \overline{\delta}(A,x) = \limsup_{h \searrow 0} \frac{\mu(A\cap(x-h,x+h))}{2h} $$ and $$ \underline{\delta}(A,x) = \liminf_{h\searrow 0} \frac{\mu(A \cap (x-h,x+h))}{2h} $$

By our standard result we have that $$\overline{\delta}(A,x) = \overline{\delta}(\hat{A},x), \qquad \underline{\delta}(A,x) = \underline{\delta}(\hat{A},x)$$

By Lebesgue differentiation theorem, we have for a.e. $x$ that $\overline{\delta}(\hat{A},x) = \underline{\delta}(\hat{A},x) \in \{0,1\}$. Hence the same holds for $A$.

So how exactly is the case different between measurable and non-measurable sets? For measurable sets $M,N$ we have $\mu(M\cap N) + \mu(M^c\cap N) = \mu(N)$. For a potentially non-measurable $A$ we only have $\mu(A \cap N) + \mu(A^c \cap N) \geq \mu(N)$. That is, the problem with non-measurability is pronounced when you take the complement.

Translated to the current set up, what we have is that for a measurable set, we have

$$ \overline{\delta}(M,x) = 1 - \underline{\delta}(M^c,x) $$

but for potentially non-measurable sets we only have

$$ 1 \leq \overline{\delta}(A,x) + \underline{\delta}(A^c,x) \leq 2 $$

In fact, one can prove the following (before you ask: the underlines are correct, we are using that $\overline{\delta} \geq \underline{\delta}$ trivially):

Theorem: the following are equivalent

  1. $A$ is not measurable
  2. There exists $x$ such that $\underline{\delta}(A,x) + \underline{\delta}(A^c,x) > 1$
  3. There exists a set $S$ of positive measure such that for every $x\in S$, we have $\underline{\delta}(A,x) + \underline{\delta}(A^c,x) = 2$.

In fact, if I am not making a mistake, the arguments in the above-linked paper prove the following result, which seems to be the closest positive answer to the question you posed.

Corollary: Let $A$ be an arbitrary set in $\mathbb{R}$. Then considering the limit $$\lim_{h\searrow 0} \frac{\mu(A\cap (x-h,x+h))}{\mu(A\cap (x-h,x+h)) + \mu(A^c \cap (x-h,x+h))}$$ we have

  1. The limit exists for almost every $x$
  2. Whenever the limit exists, it takes value in $\{0,1/2,1\}$.
  3. $A$ is measurable if and only if $1/2$ is not an output value.
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