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What's an example of a Riemann integrable, non-regulated function?

Definitions:

Let $X$ be a normed space and $[a,b]$ be a compact of $\mathbb R$.

Step Functions:

A function $f: [a,b] \longrightarrow X$ is said to be a step function if: there exists a partition $P=\{x_0 = a, x_1, \ldots, x_n = b \}$ of $[a,b]$, such that $f$ is constant on each subinterval $(x_i, x_{i+1})$.

Riemann integrable Functions:

A function $f: [a,b] \longrightarrow X$ is said to be integrable if for any $\epsilon > 0$, there are two step functions $\Phi: [a,b] \longrightarrow X$, and $\theta: [a,b] \longrightarrow \mathbb R_+$, such that:

$$\begin{cases} ||f(x) - \Phi(x)|| \le \theta(x), \ \forall \ x \in [a,b] \\ \int_a^b \theta(x)dx \le \epsilon \end{cases}$$

Regulated Functions:

$f: [a,b] \longrightarrow X$ is said to be regulated if:

$\forall \ \epsilon > 0, \ \exists \ \Phi: [a,b] \longrightarrow X$, a step function, such that: $||f(x) - \Phi(x)|| < \epsilon$, $\forall$ $x \in [a,b]$.

A regulated function is Riemann integrable: for $\epsilon > 0$, there is a step $\Phi: [a,b] \longrightarrow X$, such that: $||f(x) - \Phi(x)|| < \epsilon/4(b - a)$. We then take: $\theta(x) = \epsilon/2(b-a)$. So that $(\Phi, \theta)$ is a pair of step functions associated to $f$.

What about the converse? Need a Riemann integrable function be regulated? I am thinking not. But I'm not finding a way to prove it.

Thank you.

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    $\begingroup$ @GiuseppeNegro: I have edited the question. $\endgroup$ – user230734 May 30 '15 at 21:21
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Hint: Regulated function has left and right limit at every point.

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    $\begingroup$ Please elaborate. $\endgroup$ – user230734 May 30 '15 at 21:25
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    $\begingroup$ what should I elaborate, how to end with that hint or why it is true ? $\endgroup$ – J.E.M.S May 30 '15 at 21:37
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    $\begingroup$ Both if you may. $\endgroup$ – user230734 May 30 '15 at 21:46
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    $\begingroup$ Here is a proof books.google.pl/…, than we must produce function which does not have for example left side limit in $b$, that is for example function which is constantly 0 except for point $b-\frac{1}{n}$ where we put $(-1)^n$ $\endgroup$ – J.E.M.S May 30 '15 at 22:00
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The function $f:[0,2] \to \Bbb{R}$ $$f(x) = \left\{ \begin{matrix} 1 & \mbox{ if } x=1+\frac{1}{n} \mbox{ for some $n \in \Bbb{N}$} \\ 0 & \mbox{otherwise} \end{matrix}\right.$$ is Riemann integrable, but it is not regulated.

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    $\begingroup$ It is regulated by definition $\endgroup$ – J.E.M.S May 30 '15 at 22:04
  • $\begingroup$ @J.E.M.S it does not have right limit at 1 ? Also it looks pretty much like your example above. $\endgroup$ – Noix07 Jun 5 '17 at 19:11

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