Some classmates and I have been working through a sequence of problems in Royden's real analysis text, which are in the chapter on Lebesque measurable functions revolving around the Sequential Pointwise Limits and Simple Function approximations. We have some of them done, but are stuck on others.

For each of the problems, we assume we have $I$ a closed/bounded interval.

  1. Let $E$ be a measurable subset of $I$. Let $\epsilon > 0$. Show that there is a step function $h$ on $I$ and a measurable subset $F$ of $I$ for which $h=\chi_E$ on $F$ and $m(I\setminus F)<\epsilon$
  2. Let $\psi$ be a simple function defined on $I$. Let $\epsilon > 0$. Show that there is a step function $h$ on $I$ and a measurable subset $F$ of $I$ for which $h=\psi$ on $F$ and $m(I\setminus F)<\epsilon$
  3. Let $f$ be a bounded measurable function defined on $I$. Let $\epsilon > 0$. Show that there is a step function $h$ on $I$ and a measurable subset $F$ of $I$ for which $|h-f|<\epsilon$ on $F$ and $m(I\setminus F)<\epsilon$.

Clearly all these problems are very similar and build upon one another. You are then showing existence of a step function on $F \subset I$ where $m(I\setminus F)<\epsilon$. In the first you show that $\chi_E$ exists. Then you show a simple function $\psi$ exists which we know is of the form $\psi=\sum_{k=1}^n a_k \chi_{E_k}$. Then you do it for any bounded measurable function $f$.

Ideas for 1: There exists a finite open cover of $I$ ($O=\bigcup_{k=1}^n I_k$), which should also cover $E \subset I \subset O$ with the property that $m(O\setminus I)<\epsilon$. If we can set $F=(O\setminus E)^c$ we could try to show that $m(I\setminus F) = m(O\setminus E)<\epsilon$. Also we notice that $\chi_E = \chi_O$ on $F$. Though we haven't quite put it all together.

Ideas for 2: Let $\psi=\sum_{k=1}^n a_k \chi_{I_k}$ where again $O=\bigcup_{k=1}^n I_k$ is an open cover of $I$. We also know there is a closed $F_i \subset I_i$ where $m(I_i \setminus F_i)<\frac \epsilon n$ which could lead to $m(I \setminus F) \le m(\bigcup_{i=1}^n I_i \setminus \bigcup_{i=1}^n F_i) = \sum_{i=1}^n m(I_i \setminus F_i) < \sum_{i=1}^n \frac \epsilon n = \epsilon$

Ideas for 3: Use simple approximation theorem which says that $f$ is measurable on $E$ iff exists a sequence of simple functions which converge p.w. on $E$ to $f$ s.t. $|\psi_n|\le|f|$ on $E$ for all $n$.

Any suggestions towards putting the ideas together or simpler solutions would be greatly appreciated!


What you are missing in 1 is that you want to use your cover related to $E$ and not $I$.

Since $E$ is measurable and $E\subset I $, there exists an open cover $O=\bigcup_1^\infty I_k$ of $E$ such that $$m(O\setminus E)\leq\sum_km (I_k)-m (E)<\varepsilon/2.$$ As the sum of the series is finite, we may choose $n $ with $\sum_{k>n}m (I_k)<\varepsilon/2$. Put $O'=\bigcup_1^nI_k $. We have $$m (E\cap O')=m (E)-m (E\cap (O\setminus O')>m (E)-\varepsilon /2. $$

Let $h=\sum_1^n\chi_{J_k}=\chi_{O'}$, where $J_k=I_k\setminus \cup_1^{k-1}I_j $; each $J_k $ is a finite union of disjoint intervals, so $h $ is a step function. Let $F=(E\cap O')\cup (I\setminus O)$. Then $$ m(I\setminus F)=m(O\setminus (E\cap O'))\leq m (O)-m (E\cap O')<m (O)-m (E)+\varepsilon/2<\varepsilon. $$ And $h$ is $1$ on $E\cap O'$ and $0$ on $I\setminus O$, so $h=\chi_E$ on $F$.

For 2, remember that $\psi$ is simple, not step. So $\psi=\sum_1^na_k\chi_{E_k}$ with $E_k$ measurable. Now apply 1 to each $E_k$, to obtain $F_1,\ldots,F_n$ measurable and $h_1,\ldots,h_n$ step with $h_k=\chi_{E_k}$ on $F_k$, and $m(I\setminus F_k)<\varepsilon/n$. Let $h=\sum a_k h_k$ (sum of step is step). Let $F=\bigcap F_k$. Then $h=\psi$ on $F$, and $$ m(I\setminus F)\leq\sum m(I\setminus F_k)<\varepsilon. $$

For 3, fix $\varepsilon>0$. Since $f$ is bounded and measurable, there exists $\psi$ simple with $|f-\psi|<\varepsilon$. By 2, there exists $h$ step and a measurable set $F$ with $h=\psi$ on $F$ and $m(I\setminus F)<\varepsilon$. On $F$, $|f-h|=|f-\psi|<\varepsilon$.

  • $\begingroup$ Thanks, made it very simple, I was confused on number 1 because of the open cover! $\endgroup$ – KUSH Nov 19 '12 at 22:27
  • 1
    $\begingroup$ You are welcome. What you should ask yourself in a question like 1, is "where am I using that $E$ is measurable?" That's what made me choose the cover like I did. $\endgroup$ – Martin Argerami Nov 19 '12 at 22:33
  • $\begingroup$ why from $E$ is measurable, we have there exists open interval cover? $\endgroup$ – DuFong May 9 '16 at 20:51
  • $\begingroup$ It's how one defines the Lebesgue measure. $\endgroup$ – Martin Argerami May 9 '16 at 20:56
  • $\begingroup$ @MartinArgerami why must the union be finite? $\endgroup$ – qbert Feb 16 '17 at 0:40

I just stumbled across this post after I was trying this problem. Part one of martins incorrect but a small change can fix it. We dont know that there exists a finite open cover, ${\cup_{k=1}^n I_k}$ of $E$ such that if $O=\cup_{k=1}^n I_k$ then $m(O\setminus E) < \epsilon$ since the definintion of lebesgue measure uses countable coverings and finite coverings will not suffice. Also you assumed that $\cup_{k=1}^n I_k$ were disjoint since you said $h=\sum_1^n\chi_{I_k}=\chi_O$ but this cant be assumed using the definition of lebesgue measure. But since we know that $E \subset I $ is measurable and $m(I) < \infty $ we can use that fact that there exists a finite disjoint collection of open intervals, $\{I_j\}_{j=1}^n$, such that $m(E \setminus O) +m(O\setminus E) < \epsilon $. Then we let $F=(E\cap O) \cup \left(I \setminus (E\cup O) \right)$ and $h=\sum_{j=1}^n\chi_{I_k}= \chi_O$

  • $\begingroup$ There is no mistake, Martin just used the theorem you cited that allows one to find a finite disjoint cover of a measurable set with finite outer measure with open intervals without stating it explicitly. That there is a cover with open intervals only is due to the definition of the Lebesgue measure. $\endgroup$ – Alp Uzman Oct 18 '16 at 14:28
  • $\begingroup$ @AlpUzman That's false. Take $\mathbb{Q}$. $\endgroup$ – qbert Feb 16 '17 at 0:38
  • $\begingroup$ The cover we are talking about throughout this thread is measure theoretical, which requires the noncovered part (or the symmetric difference) to have small (or zero) measure. For $\Bbb{Q}$ take $\{]-\epsilon/2,\epsilon/2[\}$. $\endgroup$ – Alp Uzman Feb 16 '17 at 2:31
  • $\begingroup$ @AlpUzman but then you will have a problem with your definition of the step funciton no? You need the cover to be set theoretic cover of $E$ to assert that $h$ agrees with $\chi_E$ on $F$ $\endgroup$ – qbert Feb 16 '17 at 4:35
  • $\begingroup$ $F$ need not be a subset of $E$ though. $\endgroup$ – Alp Uzman Feb 16 '17 at 5:07

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