[Math] Characteristic functions based proof problem.

Question

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.

Answer 1

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 $U$.