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I am trying to show that if $T$ be a closed bounded interval and $E$ a measurable subset of $T$. Let $\epsilon >0$, then there is a step function $h$ on $T$ and a measurable subset $F$ of $T$ for which $h=\chi_E$ on $F$ and $m(T - F)<\epsilon$.

Some Thoughts towards the proof.

$\forall \epsilon >0$, there is a finite disjoint collection of open intervals $\left\{ I_{k}\right\} _{k=1}^{n}$ for which if $O=U_{k=1}^{n}I_{k}$ then $$m^{*}(E-O) + m^{*}(O-E) < \epsilon.$$

Also $$\chi_{E}=\begin{cases}1, & x\in E \\ 0, & x\notin E. \end{cases} $$

So $\forall \epsilon >0$ we have $\chi_{E} = \chi_{O}$.

Now $h$ being a step function looks like $$h =\sum _{i=0}^{n}\alpha _{i}\chi _{T_i}\left( x\right).$$

I am unsure how to proceed from here and show the two results.

Any help would be much appreciated.

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@Nate Eldredge Thanks for the formatting i wanted to learn how to do this bit. –  Hardy Aug 2 '12 at 2:53
6  
Apply Littlewood's first principle to find a finite collection of open intervals whose union $U$ satisfies $\mu(U\Delta E)<\epsilon$. Take $F$ to be $(U\Delta E)^C$ and define the step function to be $1$ on $U$ and $0$ off $U$. –  David Mitra Aug 2 '12 at 3:20
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It can be taked from here since a measurable set can be written as countable union of sets of finite measure (if the measure is $\sigma-$finite). –  leo Nov 1 '12 at 16:27

1 Answer 1

up vote 1 down vote accepted

As David Mitra suggests in the comments,

Apply Littlewood's first principle to find a finite collection of open intervals whose union $U$ satisfies $\mu(U\Delta E)<\varepsilon$. Take $F$ to be $(U\Delta E)^C$ and define the step function to be $1$ on $U$ and $0$ off $U$.

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