Let $\{R_n\}\subset (0,1)\times(0,1)\subset \mathbb R^2$ be a sequence of rectangles. For each $R_n$, we assume that the shortest side-length is at least $1/4$.
I am trying to prove that the limit of $R_n$, up to extraction of a subsequence, is a rectangle with shortest side-length at least $1/4$ too. That is, if I assume $\chi_n:=I_{R_n}$, where $I_{R_n}$ denotes the indictor function, then I am trying to prove that $\chi_n\to \chi$ where $\chi=I_{R}$ where $R$ is again a rectangle with shortest side-length is at least $1/4$.
Here is what I tried: It is clear that $$ \sup_n \|\chi_n\|_{BV(Q)}<+\infty. $$ Hence there exists a $\chi\in BV$ such that $\chi_n\to \chi$ weakly in $BV$, and hence strongly in $L^1$ and point-wisely a.e. Therefore, I know that there exists a set $S\subset Q$ such that $\chi = I_S$.
But how may I show that $S$ is a rectangle and has shortest side-length is at least $1/4$? It looks pretty obvious but I don't know how to show it mathematically...
Please advise!
You do not need such a large sledgehammer.
Denote your rectangles by $R_n = (a_n, b_n) \times (c_n, d_n)$. Then, we pick a subsequence (without relabeling) such that the sequences $\{a_n\}, \{b_n\}, \{c_n\}, \{d_n\}$ converge towards $a,b,c,d$, respectively. Then, it is easy to check that the limit in $L^1$ is the characteristic function of $R = (a,b) \times (c,d)$. Finally, the side-length condition is trivial to establish.