uniform limit of step function 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. 


*

*Prove that every continuous function on a compact interval is a uniform limit of step functions.

*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?
 A: 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.

*

*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  $f$


*You are right.
