TITLE: Can the real numbers be equally split into two sets of same measure?
QUESTION [10 upvotes]: The rational numbers $\Bbb Q$ are dense in $\Bbb R$, but they are still a set of measure zero, i.e.
$$\begin{align}
\mu(\Bbb Q \cap [a,b]) &= 0 \\
\mu((\Bbb R\!\setminus\! \Bbb Q) \cap [a,b]) &= b-a \\ \tag 1
\end{align}$$
for any finite interval $[a,b]$.
Is it possible to have more equally distributed sets, so that neither of them is a set of measure 0, and this holds on any interval similar to (1)?
More specifically, are there decompositions $A, B\subset \Bbb R$ and a measure $\mu$ such that all of the following conditions hold?
$$\begin{align}
A\cap B = \emptyset\quad&\text{ and }\quad A\cup B = \Bbb R \\
\mu ([a,b]) &= b-a \\
\mu (A\cap [a,b]) &= (b-a) / 2 \\
\mu (B\cap [a,b]) &= (b-a) / 2 \\\tag 2
\end{align}$$
for any finite interval $[a,b]\subset\Bbb R$? The first line just states that $A,B$ is a decomposition of $\Bbb R$, the second line is a common normalizing condition for $\mu$.
Or, at your option, that
$$\begin{align}
\mu(A\cap[a,b]) &= (b-a)\kappa \qquad\text{ for some } 0<\kappa<1 \\
\mu(B\cap[a,b]) &= (b-a)(1-\kappa)
\end{align}$$
again for any finite interval $[a,b]$.  And it might even be in order if $\kappa=\kappa(a,b)$ depends on $a$ and $b$ provided $0<\kappa(a,b)<1$ for finite intervals.
My intuition says that there is no such decomposition, but maybe I am wrong.

REPLY [7 votes]: No.
First, it's clear that $\mu([a,b])=b-a$, and hence that $\mu$ is just Lebesgue measure. Now the Lebesgue Differentiation Theorem (applied to $\chi_A$) shows that $$\frac12=\lim_{h\to0}\frac{\mu(A\cap[a-h,a+h])}{2h}=1$$for almost every $a\in A$, contradiction.