Given an open interval $I=(a, b)\subset \mathbb R$ for some $a, b\in \mathbb R$ and a countable union of its open subintervals $U=\bigcup\limits_{i=1}^\infty J_i$ such that $\sum\limits_{i=1}^\infty L(J_i)<\infty$($L(J_i)$ is the length of the interval $J_i$), for any $\epsilon>0$ can we choose an finite union of open subintervals $K=\bigcup_{i=1}^n K_i\subset I$ and an integer $N$ such that $$\bigcup\limits_{i=N+1}^\infty J_i\subset K\text{ and } \left| L(K)-\sum_{i=N+1}^\infty L(J_i)\right|<\epsilon?$$

I came up with this question when I was attempting to prove a problem in real analysis, and this would finish the proof if it is true.

  • $\begingroup$ Can you assume the $J_i$ are pairwise disjoint? (Otherwise $\sum L(J_i)$ could be infinite.) $\endgroup$ – David Mitra Apr 20 at 8:02
  • $\begingroup$ I can assume that the sum $\sum L(J_i)<\infty$. $\endgroup$ – William Sun Apr 20 at 8:06
  • $\begingroup$ Even if the $J_i$ were pairwise disjoint, I don't think this is possible. Take $L_1$, $L_2$, $\ldots$ to be disjoint open intervals in $(0,1/8)$ and $R_1$, $R_2$, $\ldots$ to be disjoint open intervals in $(7/8,1)$. Take $(J_i)=R_1,L_1,R_2,L_2,\ldots$. Then $|L(K)-\sum_{i={N+1}}^\infty L(J_i)|$ is always at least $1/2$. $\endgroup$ – David Mitra Apr 20 at 8:10
  • $\begingroup$ We can assume that the set $K$ is a finite union of open intervals instead of a single one. Sorry it takes efforts to examine the least condition I need in proof and I just found out. $\endgroup$ – William Sun Apr 20 at 8:16
  • $\begingroup$ Where is $U$ used again? Are the $J_i$ contained in $I$? $\endgroup$ – zhw. Apr 20 at 18:43

Let $(q_i)$ be an enumeration of the rationals in $I$. Any such enumeration has the property that the tails of $(q_i)$ are always dense in $I$. (Any point of $I$ is the limit of a sequence of distinct rationals - a subsequence of $(q_i)$. But a tail of $(q_i)$ will contain a tail of the subsequence.)

Let $\epsilon > 0$ and let $$J_i = (q_i - \epsilon 2^{-i-1}, q_i + \epsilon 2^{-i-1})\cap I$$ Then $L(J_i) \le \epsilon 2^{-i}$, so for any $N > 0,$ $$\sum_{i>N} L(J_i) \le \sum_{i>N} \epsilon 2^{-i} = \epsilon 2^{-N} < \epsilon$$

By choosing $\epsilon$ small enough, we can make $\sum_{i} L(J_i)$ as small as we please. In particular, make it smaller than $\frac{b-a}2$.

Because $(q_i)_{i > N} \subseteq \bigcup_{i>N}J_i$ is dense in $I$, any finite union of open intervals $K_N \subseteq I$ containing $\bigcup_{i>N}J_i$ can miss at most a finite number of points of $I$, so $L(K_N) = b-a$, a fixed value with respect to $N$. Therefore, for all $N$,$$\left| L(K_N)-\sum_{i>N} L(J_i)\right| > \frac {b-a}2$$


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