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The Riemann function $f(x)=\begin{cases}\frac 1 q \text{ if } x=\frac p q \in\Bbb Q\\ 0\text{ if } x\in \mathbb R\setminus \Bbb Q\end{cases}$

is a step function. Then, for any epsilon construct a step function $k:[0,1]\to \mathbb R$ such that $||f-k||=\sup\{|f(t)-k(t)|:t \in [0,1]\}<\epsilon$

Suggestion: restrict to $q<1/\epsilon$

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    The Riemann function is a step function? If it were, you could just take $k$ to be $f$. Anyway, it's not good to just dump questions here without telling us what you know about the question, what you have tried, where you get stuck, and so on.2012-10-24

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Here is a suggestion to get you started : define a function $t_\varepsilon$ by $t_\varepsilon(x)=x\mathbf 1_{x\geqslant\varepsilon}$. Then, $k=t_\varepsilon\circ f$ is a step function (why?) such that $f-\varepsilon\lt k\leqslant f$ (why?).

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    To which only the silence of the abysses responded...2012-10-31