I'm trying to prove the next proposition:
Suppose that $f\in C_K$ has support in the compact interval $[a, b].$ Given any dense subset $A$ of $\mathbb{R}$ and $\epsilon > 0,$ there exists an $A-$valued step function $f_{\epsilon}$ on $(a, b)$ such that $$\sup_{x\in\mathbb{R}}|f(x)-f_{\epsilon}(x)|\leq\epsilon.$$
I'm stuck. I was thinking in use Stone-Weiestrass Theorem, so there is a sequence of polynomials $\{P_{n}\}$ such that $P_{n}(x)\rightarrow f(x) uniformly.$ I'd like to built a $A-$step function utilizing the convergence of polynomials such that $\sup_{x\in\mathbb{R}}|f(x)-f_{\epsilon}(x)|\leq\epsilon,$ but I don't get it.
Any kind of help is thanked in advanced.
Let $f\colon [a,b]\to \mathbb R$, $A\subset \mathbb R$ and $\varepsilon>0$ be as above. Since $f$ is defined on a compact, it is uniformly continuous, so we can find a $\delta>0$ such that for all $x,y\in[a,b]$ we have $$|x-y|\leq\delta \Rightarrow |f(x)-f(y)|<\frac \varepsilon2.$$
Now subdivide the interval $[a,b]$ into pieces $$[a,a+\delta)\cup [a+\delta, a+2\delta)\cup [a+2\delta,a+3\delta) \cup \dots \cup [a+n\delta, b],$$ where $n$ is the largest natural number with $a+n\delta\geq b$.
Now we define $f_\varepsilon$ to be constant on each piece $I_i:=[a+i\delta,a+(i+1)\delta)$ with value $\alpha_i\in A$ chosen such that $|f(a+i\delta)-\alpha_i|<\frac\varepsilon 2$ (this exists because A is dense in $\mathbb R$).
Now it is easy to calculate that for every $x\in I_i$ we have (using $|x-(a+i\delta)|\leq \delta$): $$|f(x)-f_\varepsilon(x)|=|f(x)-f(\alpha_i)|<|f(x)-f(a+i\delta)|+|f(a+i\delta)-\alpha_i|< \frac\varepsilon 2+\frac\varepsilon 2=\epsilon.$$
Hence we conclude that $$\sup_{x\in [a,b]}|f(x)-f_{\epsilon}(x)|<\epsilon.$$