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Suppose we have $E\subseteq [a,b]$ is a measurable set. How do we prove that for all $\epsilon>0$, there exists finitely many disjoint closed intervals $\{[a_i,b_i]\}_{i=1}^N$ in $[a,b]$ such that $$|\{x\in [a,b]:\chi_E\neq\chi_{\bigcup[a_i,b_i]}\}|<\epsilon$$?


I have made an attempt, not sure if it is ok:

Let $\epsilon>0$. Write $E=\bigcup E_k$, where $E_k$ are disjoint measurable components of $E$. Since $E\subseteq [a,b]$, only finitely many $E_k$ has nonzero measure say $|E_1|,\dots, |E_N|>0$.

For $1\leq k\leq N$, since $E_k$ is measurable, there exists a closed set $F_k\subseteq E_k$ such that $|E_k\setminus F_k|<\epsilon/N$.

Let $[a_k, b_k]$ be the smallest closed interval containing $F_k$. Then $|E_k\setminus [a_k, b_k]|\leq|E_k\setminus F_k|<\epsilon<N$.

So $$|\{x\in [a,b]:\chi_E\neq\chi_{\bigcup[a_i,b_i]}\}|=\sum_{k=1}^N|\{x\in E_k: \chi_{E_k}\neq\chi_{[a_k, b_k]}\}|<\epsilon/N\cdot N=\epsilon$$

It is clear that $[a_k, b_k]$ are disjoint since $F_k$ are disjoint.

Anything wrong here? Thanks.

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I don't understand why finitely many of the $E_k$ should have non-zero measure. Take a look at $$E=\left[\dfrac{1}{2},1\right]\cup\left[\dfrac{1}{4},\dfrac{3}{8}\right]\cup\left[\dfrac{1}{8},\dfrac{3}{16}\right]\cup\cdots$$ for instance. All you can say is that the series $$\sum_{k\geq 1}|E_k|$$ is convergent.

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  • $\begingroup$ Oh yes, you are right. Any hint on the correct approach? $\endgroup$ – yoyostein Sep 28 '16 at 10:57
  • $\begingroup$ Here is a good answer: math.stackexchange.com/questions/216579/…. It doesn't really matter whether the intervals are open or closed since it doesn't change their measure. $\endgroup$ – Olivier Moschetta Sep 28 '16 at 11:04
  • $\begingroup$ Thanks. If we used closed intervals they would not be disjoint but just non-overlapping, is that correct? $\endgroup$ – yoyostein Sep 28 '16 at 11:24
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This is my attempted answer (based on the reference given by Olivier). Any comments are welcome.

Let $\epsilon>0$. Since $E$ is measurable, there exists an open set $G\supseteq E$ such that $|G\setminus E|<\epsilon/2$.

Since $G$ is open, $G=\bigcup_{k=1}^\infty I_k$ where $I_k$ are disjoint open intervals.

Since $|G|=|E|+|G\setminus E|<\infty$, so $\sum_{k=1}^\infty |I_k|<\infty$. Thus there exists $N$ such that $\sum_{k=N+1}^\infty |I_k|<\epsilon/4$.

For each $1\leq k\leq N$, choose a closed interval $[a_k, b_k]\subseteq I_k$ such that $$|(\bigcup_{k=1}^N I_k)\setminus(\bigcup_{k=1}^N [a_k, b_k])|<\epsilon/4.$$

Then \begin{align*} |\{x\in [a,b]:\chi_E\neq\chi_{\bigcup [a_k, b_k]}\}|&=|E\triangle(\bigcup_{k=1}^N [a_k, b_k])|\\ &=|E\setminus(\bigcup_{k=1}^N [a_k, b_k])|+|(\bigcup_{k=1}^N [a_k, b_k])\setminus E|\\ &\leq|G\setminus (\bigcup_{k=1}^N [a_k, b_k])|+|G\setminus E|\\ &\leq|G\setminus(\bigcup_{k=1}^N I_k)|+|(\bigcup_{k=1}^N I_k)\setminus(\bigcup_{k=1}^N [a_k, b_k])|+\epsilon/2\\ &<\epsilon/4+\epsilon/4+\epsilon/2=\epsilon. \end{align*}

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