Lebesgue density of a union of closed disks. Let $E$ be the union of uncountably many closed disks in $\mathbb{R}^2$ with radii at least $1.$ If we let $\partial E$ denote the boundary of $E$ and $x \in \partial E,$ is $$\lim_{r \to 0} \frac{m((B(x,r)\cap \partial E) }{m(B(x,r))} <1$$ whenever the limit exists? 
Here $m$ is the Lebesgue measure and $B(x,r)$ is the ball of radius $r$ centered at $x$. 
 A: Let $D(x,r)$ denote the closed disc with center $x$ and radius $r.$ Let $E = \cup_x D(x,1),$ where $x$ runs through some set $X.$ Area measure will be denoted by $A.$
The main idea is this: No point $x_0\in \partial E$ can lie in the interior of any $D(x,1)$ with $x\in X.$ However, the boundaries of such discs must get close to $x_0.$ So fix an $r<1.$ Draw pictures. Doesn't it look like likely that the interiors of such discs will, in area, contain a fixed percentage of $D(x_0,r)?$ Answer: Yes! And this implies $\partial E$ will be missing a fixed percentage of $D(x_0,r)$ for any $r<1.$
Thus the following result seems likely: There exists $c\in (0,1)$ such that
$$\tag 1 \limsup_{r\to 0} \frac{A(\partial E \cap D(x_0,r))}{A(D(x_0,r))} \le c$$
for all $x_0 \in \partial E.$
Note that if we prove this, we've actually shown $A(\partial E) = 0.$ That's because if $A(\partial E) > 0,$ then for a.e. $x_0 \in \partial E$ we would have the limit of the fraction in $(1)$ equal to $1.$ That follows from the Lebesgue point of density theorem. But $(1)$ rules this out.
So how do we verify $(1)?$ It's better to suggest some steps rather than going through all the details. First, by scaling, we see that
$$\tag 2 \frac{A(D((0,0),r)\cap D((r,0),r))}{A(D((0,0),r))}$$
is independent of $r>0.$ Thus $(2)$ is a constant $b\in (0,1).$ Next, note that for fixed $r>0,$
$$\lim_{\epsilon\to 0^+} \frac{A(D((0,0),r)\cap D((r+\epsilon,0),r))}{A(D((0,0),r))} = b.$$
Finally, note that if $r<1,$ then $D(r+\epsilon,r)\subset D(1+\epsilon,1).$ Thus, if $r<1,$
$$\lim_{\epsilon\to 0^+} \frac{A(D((0,0),r)\cap D(1+\epsilon,1)}{A(D((0,0),r))} \ge b.$$
Use these results, the rotational symmetry available, and the preliminary remarks to see that $(1)$ holds, with $c= 1-b.$
