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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 $h=X_E$ 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 $X_E$ exists. Then you show a simple function $\psi$ exists which we know is of the form $\psi=\sum_{k=1}^n a_k X_{E_k}$. 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 $X_E = X_O$ on $F$. Though we haven't quite put it all together.

Ideas for 2: Let $\psi=\sum_{k=1}^n a_k X_{I_k}$ 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!

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1 Answer 1

up vote 2 down vote accepted

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

Since $E$ is measurable, there exists an open cover $O=\bigcup_1^n I_k$ of $E$ such that $m(O\setminus E)<\varepsilon$. Let $h=\sum_1^n\chi_{I_k}=\chi_O$. Let $F=E\cup (I\setminus O)$. Then $$ m(I\setminus F)=m(O\setminus E)<\varepsilon. $$ And $h$ is $1$ on $E$ 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 $F$, $|f-h|=|f-\psi|<\varepsilon$.

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Thanks, made it very simple, I was confused on number 1 because of the open cover! –  KUSH Nov 19 '12 at 22:27
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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. –  Martin Argerami Nov 19 '12 at 22:33
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