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?
Best Answer
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.