function $f:[0,1] \to \Bbb R$ s.t $f$ is not a step function, but for any $\epsilon \in (0,1)$ the restriction of $f$ to $[\epsilon,1]$ is.

Question

I need to construct a function $f:[0,1] \to \Bbb R$ such that $f$ is not a step function, but for any $\epsilon \in (0,1)$ the restriction of $f$ to $[\epsilon,1]$ is a step function.

So my function would be:

$f(x)=\begin{cases}x^2, \text{ for $\in[0,1-\epsilon]$}\\f_i(x) , \text{ for $x\in(p_{n-1},p_n) \subset P_{i} \subset (1-\epsilon,1]$}\end{cases}$

Where $P_i=\{p_{i_0},p_{i_1},...,p_{i_k}\}\in\{P_1,P_2,...,P_k\}$ is a partition compatible with $f_i(x)$ on the subset $(1-\epsilon,1]$

Would this make any sense? I want the partitions to basically be dependent on the choice of $\epsilon$ we make, so that's why we have a set of partitions

Answer 1

Consider the function $$f(x) = \begin{cases} \chi({\sin(1/x)}), \ &x\in(0,1] \\ 0, \ &x = 0 \end{cases}$$ where $$\chi({x}) = \begin{cases} 1, \ &x\geq0 \\ 0, \ &x< 0\end{cases}$$ Then on any interval $[\epsilon,1]$ with $\epsilon\in(0,1)$, the restriction $f|_{[\epsilon,1]}$ is a finite sum of characteristic functions and hence a step function. But on $[0,1]$ the sum is infinite and is therefore not a step function.