Measure of intersection of set and its translation I came across an old qualifying exam question:
Let $A\subset [0,1)$ be a Lebesgue measurable subset of unit intreval such that $0<\mu(A)<1$. For every $x\in [0,1)$ let $A+x=\{x+y$ mod 1$:y\in A\}$. Prove that there exists $x_0=x_0(A)\in[0,1)$ such that $\mu(A\cap(A+x_0))<\mu(A)$
Intuitively, it seems that if $A$ is a measurable interval then there exist an interval $I$ such that they overlap by more than half. Moreover the intersection $(A\cap(A+x_0))$ would need less intervals to "cover up" than $A$. Hence we have the inequality. But I am having a hard time making this rigorous.
 A: \begin{equation}
\begin{split}
\int_0^1 \mu(A\cap A+x)dx &= \int_0^1 \int_A \chi_{A+x}\ (y)dydx \\
&= \int_0^1\int_A \chi_A(y-x) dy dx\\
&= \int_A\int_0^1 \chi_A(y-x) dx dy \ \ \ \ \ \ \text{(Fubini's theorem)}\\
&=\int_A \mu(A) dx = \mu(A)^2 <\mu(A)
\end{split}
\end{equation}
as $0< \mu(A) <1$. Thus there is $x_0\in [0,1]$ so that 
$$\mu(A \cap A+x_0) < \mu(A). $$
A: Almost every point of $A$ has Lebesgue's density $1$. By
Lebesgue's density theorem, the density of $A$ is $0$ or $1$ at almost every point in $[0,1]$. Also, almost every point of $A$ has density $1$. More precisely the symmetric difference of $A$ and the set of points of density $1$ for $A$ has measure $0$. 
Take any point $x\in[0,1)$ such that $A$ has density $0$ at $x$, and any point $y$ such that $A$ has density $1$ at $y$. Let $x_0=y-x$. Then it is easy to show that $\mu(A\cap(A+x_0))<\mu(A)$. 
The density of a set $A$ at a point $x$ is defined as the limit 
$\lim_{\varepsilon\to0}\dfrac{A\cap(x-\varepsilon,x+\varepsilon)}{2\varepsilon}$. 
The idea, as I see it, is that even if $A$ is dense in $[0,1]$ (in the topological sense), there are nevertheless some points $x$ near which $A$ is "small" (and so the complement of $A$ is big), and other points $y$ near which $A$ is "big". When we translate $A$ so that $x$ goes to $y$, then the intersection $A\cap([0,1]\setminus(A+y-x))$ is "big" (has density $1$ at $y$), hence when we intersect $A\cap(A+y-x)$ we do throw away something "essential" (or positive measure, namely we throw out $A\cap([0,1]\setminus(A+y-x))$), so we end up with a smaller measure in the intersection, than the measure of $A$. 
