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I want to show that the indicator function (aka. the characteristic function) is not a regulated function.

\begin{align} \chi : \begin{cases}[a,b] & \longrightarrow \mathbb{R} \\ x & \longmapsto \begin{cases} 0, \ x \in \mathbb{Q} \cap [a,b] \\ 1 , \ x \in (\mathbb{R} \setminus \mathbb{Q} ) \cap [a,b] \end{cases} \end{cases} \end{align}

The definition (Königsberger Analysis) uses seems to differ to other sources where such topics are discussed.

Def: A function $f: [a,b] \to \mathbb{R}$ is called a regulated function if $\exists (f_n)_{n \in \mathbb{N}}$ sequence of step functions $f_n: [a,b] \to \mathbb{R}$ such that $$ \lim_{n \to \infty}\|f_n-f \| =0 $$ where $\|. \|$ denotes the supremum norm.

My approach: I am having trouble with the 'elegant' formulation of the solution. Also I am not sure how to do the calculation.

I want to show that for all step functions $\phi: [a,b] \to \mathbb{R}$ it follows that $\| \phi -f \| \geq 1$ or some other small value.

So let $\phi : [a,b] \to \mathbb{R}$ be a step function and $a=x_0< x_1< \dots < x_n =b$ a partition of $[a,b]$, since $\mathbb{Q}$ is dense in $\mathbb{R}$ I can always find $v_1 \in (a, x_1)\cap \mathbb{Q}$ and $v_2 \in (a,x_1) \cap (\mathbb{R} \setminus \mathbb{Q})$. Since $\phi$ is a step function it would also follow that $\phi(v_1) = \phi (v_2) = c_0 \in \mathbb{R}$

Now I want to use this for a calculation with the norm $$\| \phi - \chi \| = \sup_{x \in [a,b]} |\phi (x) -\chi(x) | \geq \max_{x \in [a,b]} | \phi (x) -\chi(x) | $$

and now, sadly, I lack experience on how to argument with the maximum or the minimum. I tried to visalize it for easier functions than $\chi$ but I couldn't see how I could split up the maximum in order to make even use of my calculations above.

I'd appreciate some insight on how to continue with my calculations, i.e. since I have partition of my interval, I can maybe split up the maximum for all these cases and always end up with states such as: $$ \geq \max(|\phi(v_1)-\chi(v_1)|, \dots , |\phi(v_n)-\chi(v_n)| \\ \geq \max(|\phi(v_1)-\chi(v_1)|, | \phi(v_2)- \chi(v_2)|) = \max (|c_0|, |c_0-1|) $$ but that just doesn't seem right to me.

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  • $\begingroup$ Use that any $y\in \Bbb{R}$ fulfils $\max \{|y|, |y-1|\}\geq 1/2$ to show $\| \chi- \phi \|\geq 1/2$. With the value $1$ you will not have success (consider $\phi \equiv 1/2$) $\endgroup$ – PhoemueX Dec 7 '14 at 8:47
  • $\begingroup$ Thanks for your comment @PhoemueX, so it looks like I am only one step away from the solution. I understand what you've said, but I still don't know how to correctly write that. Wlog I choose just one point with said characteristics and obtain that that $\| \phi - \chi \| \geq \max{(|c_y|,|c_y-1|)}\geq 1/2$ ? $\endgroup$ – Spaced Dec 7 '14 at 12:38
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    $\begingroup$ I posted a continuiation of the beginning your proof as an answer. Does that help you? $\endgroup$ – PhoemueX Dec 7 '14 at 18:42
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Your proof is nearly perfect. Simply note that

$$ \Vert \phi - \chi \Vert \geq \max \{ |\phi(v_1) - \chi(v_1)|, |\phi(v_2) - \chi(v_2)|\}\\ = \max\{ |c_0 - 0|, |c_0 - 1|\} \geq 1/2. $$

Here, we used that $v_1 \in \Bbb{Q}$ and hence $\chi(v_1) = 0$ and $v_2 \in \Bbb{R}\setminus\Bbb{Q}$, hence $\chi(v_2) = 1$ and we also used $\phi(v_1) = \phi(v_2) = c_0$ (as you already noted yourself).

Finally, we made use of the inequality

$$ \max\{|y|, |y - 1|\} \geq 1/2, $$ which is valid for all $y \in \Bbb{R}$.

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