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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!

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    Suppose that $R_1$ is a square of side $1/4$ in the upper left quadrant of $(0,1) \times (0, 1)$, $R_2$ is the same in the lower left, $R_3$ is the same in the lower right, $R_4$ is the same in the upper right, and then the sequence repeats. Does the sequence of associated characteristic functions even have a limit? – 2017-01-18
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    @JohnHughes I mean up to a subsequence. Sorry about that. I already corrected my post. – 2017-01-18

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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.

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    I understood the side-length condition is trivial, however, how may I check the limit in $L^1$ is the characteristic function of a rectangle? I know it is should foolish but could you extend your answer about this fact? thank you! – 2017-01-19
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    You can directly compute $\int |\chi_R - \chi_{R_n}|$ since everything consists of rectangles. – 2017-01-19
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You are talking about two sequence, let's call them $a_n$ and $b_n$, which represent the sides of your rectangles. You are assuming also that:

$$a_n \in (0, 1), b_n \in (0,1) ~\text{and}~ \min\{a_n, b_n\} \geq 0.25.$$

Moreover, according to your words,

I am trying to prove that the limit of $R_n$, up to extraction of a subsequence...

you assume that the limit of $R_n$ exists for any subsequence you can imagine. That is, the limit of $R_n$ exists, and then also the limit of $a_n$ and $b_n$ exist. Let's call $a$ and $b$ the limit values of the sequences $a_n$ and $b_n$. For sure, your limit rectangle $R$ is defined by $a$ and $b$. Furthermore, using the definition, $$\min\{a, b\} \geq 0.25.$$

Then, I think you are done...

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    The problem is... why the limit of $R_n$ is a rectangle? – 2017-01-18
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    Apparently, $\{R_n\}\subset (0,1)\times(0,1)$ suggests that $\{R_n\} = \{(a_n, b_n)\}$... – 2017-01-18
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    I mean, yes, each $R_n$ is a rectangle. But why the limit of $R_n$, in the $BV$ or $L^1$ sense, is still a rectangle? – 2017-01-18
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    I don't catch why $\{R_n\} = \{(a_n, b_n)\} \to \{R\} = \{(a, b)\}$ can't be a rectangle... – 2017-01-19