uniform limit of step function

Question

Define a step function to be a function that is piecewise constant, $$ f(x)=\sum_{i=1}^{n}c_i\chi_{[a_i,b_i)},$$ where $[a_i,b_i)$ are disjoint intervals.

1. Prove that every continuous function on a compact interval is a uniform limit of step functions.
2. Prove that a uniform limit of step functions (on a compact interval) is Riemann integrable.

For second one, if 1. holds, uniform convergence can be used to establish integrability and dif­ferentiability of limits of functions and the interchange of the operation and the limit. So, if $f_n$ are Riemann integrable on $[a,b]$ and $f_n$ converges uniformly to $f:[a,b]\rightarrow\mathbb{R}$, then $f$ is Riemann integrable and $$ \int_a^bf_ndx\longrightarrow\int_a^bfdx. $$ But, how can I prove the first lemma?

Answer 1

Some authors called such a function $f : I\subseteq\mathbb{R}\rightarrow\mathbb{R}$ admissible, i.e. if it is the uniform limit of step functions.

1. Let $\varepsilon=\dfrac{1}{n},$ $n\in\mathbb{N^*}$. For $\varepsilon$ we determine a $\delta>0$ such that for all $x,y\in\left[a,b\right] $ with $$ \left\vert x-y\right\vert <\delta, $$ we have $$ \left\vert f\left( x\right) -f\left( y\right) \right\vert <\varepsilon. $$ Let $m$ m the greatest integer with $$ \delta<\dfrac{b-a}{m}\text{ or }a+m\delta<b. $$ For every integer $l$ with $0\leq l\leq m$ we set $$ x_{l}:=a+l\delta\text{ and }x_{m+1}:=b $$ and define a step function by $$ f_{n}\left( x\right) =\left\{ \begin{array} [c]{l}% f\left( x_{l}\right) ,\text{ for }x_{l}\leq x<x_{l+1}\\ f\left( b\right) ,\text{ for }x=b \end{array} .\right. $$ Then for all $x\in\left[a,b\right] $ we have \begin{align*} \left\vert f\left( x\right) -f_{n}\left( x\right) \right\vert & =\left\vert f\left( x\right) -f\left( x_{l}\right) \right\vert \\ & <\dfrac{1}{n}, \end{align*} as $\left\vert x-x_{l}\right\vert <\delta$ . Finally, $\left( f_{n}\right) _{n\in\mathbb{N}}$ converges uniformly to $f$

2. You are right.