The greedy circle packing of a bounded region in the plane is the result of placing at each stage the largest possible disk into the region that remains uncovered.
The greedy circle packing of a square, for example, begins by placing the inscribed circle, and then four more circles, one in each corner, and then eight more circles adjacent to them in the corners, and so forth. It is not difficult to prove that the total measure of these circles in the limit is equal to the measure of the original square.
One way to see this is to observe that at each stage, we are packing circles into the triangular-like region between three other circles (or the straight lines of the square boundary).
The key observation is that we can always cover at least a fixed nonzero proportion $r$ of such a triangular-like region by placing finitely many circles, at least 1%, say. This is enough to imply that in the limit we achieve full measure, since if it were less, then at some finite stage we would be close enough to that value so that proportion $r$ of what remains would bring us over the supposed limit. So the packing limit must be full measure.
This argument works more generally with many other figures. My question is whether it works with every open set.
Question. Must the greedy circle packing of a bounded open set $U$ exhaust the measure of $U$?
The greedy aspect is required, since it is easy to find non-greedy but maximal circle packings of arbitrarily small measure—simply start by placing a circle of area at most $\varepsilon/2$, then another of size at most $\varepsilon/4$, and so forth, in such a way that the resulting disks are dense in the region. You can't add any more circles, so the packing is maximal, but the total measure is at most $\varepsilon$.
Perhaps one can make a counterexample to the question by starting with a square, and then strategically omitting points here and there so as to form obstacles to the circles that might be placed. For example, if you remove the center point of the square, then the largest circle that fits is much smaller than the full inscribed circle. By subsequently removing more points, one can make the next greedy circle much smaller still.
But I haven't been able to make this idea work while controlling the measure of the closure of the omitted points. For example, if the closure of the omitted points included the boundaries of all the placed circles, we would be in the fat Cantor set type situation, and the circles might still exhaust the measure of the region. Perhaps this always happens, and this way of thinking might turn into a proof of the positive result. I am unsure.
In one dimension, the answer is affirmative, since the bounded open set simply is the greedy circle packing of it. Perhaps the question is interesting in higher dimensions. (Thanks to Twitter user lemma Meringue.)
(Note: Please help me retag, since this question isn't in my usual area and I am unsure which tags are best. Also, apologies if the question turns out to be much easier than I have realized.)
Best Answer
Yes, the reason is the same as in the proof of Vitali covering theorem. It suffices that $U$ has finite measure, allowed to be unbounded.
If $D_i$ are the greedy-chosen (closed, but this is not crucial of course) disks with centers $p_i$ and radii $r_i$, let $U_i$ be the disks with centers $p_i$ and radii $2r_i$.
Note that the sum of measures of $D_i$'s is finite, since they are disjoint and $U$ has finite measure. It yields that $\lim r_i=0$ and also that the the sum of measures of $U_i$ is finite.
Lemma. Every point $x$ which is not covered by $D_i$'s lies in the union of $U_k$ for $k>N$ (for every $N$).
Proof. Since $x\notin \cup_{j\leqslant N} D_j$ and this set is closed, there exists a disk $V\subset U$ with some radius $r>0$ centered in $x$ and not intersecting with $\cup_{j\leqslant N} D_j$. Since $\lim r_i=0$, there exists minimal $m$ for which $r_m<r$. By greedy procedure, since we did not choose $V$ or larger disk on the $m$-th step, $V$ has non-empty intersection with some $D_j$, $j<m$. Thus by the choice of $V$ we get $j>N$, and by minimality of $m$ we get $r_j\geqslant r$. Therefore $x\in U_j$.
It remains to note that since the sum of measures of $U_i$ is finite, the tail sum over $i>N$ may be made arbitrarily small if $N$ is large.