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I am trying to follow a proof that characterizes regulated functions as those functions $f$ for which there exists a sequence of step functions converging uniformly to $f$. An excerpt of the text I'm using is provided below:

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Note that "jump continuous" and "regulated" are synonymous and by $f \in \mathcal{S}(I,E)$ the author just means a regulated function defined on the compact interval $I = [\alpha, \beta]$

I follow all of this proof except for the part where the refined partition $Ʒ_1$ is selected. I don't know why the first partition $Ʒ_0$ doesn't work and I don't know what the motivation is for refining it in the first place. Can anyone explain the purpose of the refined partition and why it is needed?

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I believe the original partition could also have been used; it would just have required a little extra argument. The problem that the refinement is intended to fix is that the interval $(\alpha(x_j),\beta(x_j))$ doesn't necessarily reach up to $x_{j-1}$ and $x_{j+1}$, so the construction as stated wouldn't work. I see three ways of fixing that, and they all rely on the fact that adjacent intervals $(\alpha(x_j),\beta(x_j))$ and $(\alpha(x_{j+1}),\beta(x_{j+1}))$ overlap. First, the refinement as in the text, which works because the points used for the refinement can be chosen in the overlap to ensure $\lVert f(s)-f(t)\rVert\lt1/n$ in each part of the partition; second, to use $f(\zeta_j)$ instead of $f((\xi_{j-1}+\xi_j)/2)$, where $\zeta_j$ lies in the overlap; or third, to argue that for the function as defined in the text $\lVert f(s)-f(t)\rVert\lt2/n$ by using the triangle inequality with an intermediate point in the overlap.

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    Perhaps an even more basic question, but why do the intervals necessarily overlap? I believe it is because, for example, if we consider the intervals $(\alpha(x_1), \beta(x_1))$ and $(\alpha(x_2), \beta(x_2))$ about two points $x_1 < x_2$ our "cover" will miss all of the points between $\beta(x_1)$ and $\alpha(x_2)$. If even one such instance occurs then the whole cover is blown, so to speak, and can never fully cover $I$. Therefore, all intervals must overlap. Is this the right way to think about this?2012-03-15
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    @ItsNotObvious: That's exactly the right way to think about it; and the cover being blown is a nice pun :-). Note that this is because it's an *open* cover, and two open intervals necessarily either have an overlap or a gap. We might have $\beta(x_1)=\alpha(x_2)$; then the gap would consist of that single point.2012-03-15
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    I understand this now; thanks for your help.2012-03-15
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    @ItsNotObvious: You're welcome.2012-03-15