I actually posted this question before, but didn't get a correct answer. Two distinct definitions follow; ($f$ is assumed to be a real-valued function and $\alpha$ is assumed to be a monotonically increasing function on $[a,b]$)

Definition 1: $\forall \epsilon>0, \exists \text{ a partition } P=\{x_0,...,x_n\} \text{ of } [a,b] \text{ such that } \sum_{i=1}^n [M_i - m_i] \Delta \alpha_i <\epsilon$ where $\Delta \alpha_i=\alpha(x_i)-\alpha(x_{i-1})$ and $m_i=\inf_{x\in[x_{i-1},x_i]} f(x)$ and $M_i=\sup_{x\in[x_{i-1},x_i]} f(x)$

Definition 2: $\forall \epsilon>0, \exists \text{ a partition } Q=\{x_0,...,x_n\} \text{ of } [a,b] \text{ such that } \sum_{i=1}^n [M_i - m_i] \Delta \alpha_i <\epsilon$ and $d(x_i,x_{i-1})$ is a constant where $\Delta \alpha_i=\alpha(x_i)-\alpha(x_{i-1})$ and $m_i=\inf_{x\in[x_{i-1},x_i]} f(x)$ and $M_i=\sup_{x\in[x_{i-1},x_i]} f(x)$

Obviously Definition2 implies Definition 1, but what about the converse?


Let $[a,b]=[0,1]$. $$\alpha(x) = \begin{cases} 1 \mbox{ if } x \leq \frac{1}{\sqrt{2}},\\ 0 \mbox{ otherwise.} \end{cases}$$

$$f(x) = \begin{cases} 1 \mbox{ if } x \geq \frac{1}{\sqrt{2}},\\ 0 \mbox{ otherwise.} \end{cases}$$

For the first definition consider the partition $\{0,\frac{1}{\sqrt{2}},1\}$, the relevant Darboux sum is 0. So its integrable.

For the second definition, the point $\frac{1}{\sqrt{2}}$ will be in the interior of an interval in the parition, as all $x_i$'s will have to rational as they are equally spaced, so the second kind of Darboux sum will be at least 1 irrespective of the equally spaced partition chosen, so the integral does not exist by the second definition.

  • $\begingroup$ The partition $\{0,\frac{1}{\sqrt{2}},1\}$ is equally spaced. Second definition doesn't mean that all $x_i$'s should be rationals. $\endgroup$ – Katlus Jan 24 '13 at 16:14
  • $\begingroup$ I thought the second definition implied $x_i - x_{i-1} $ is the same. Which means each partition width should be $1/k$ as the length adds upto 1, hence and $x_i = 1/k$. $\endgroup$ – Arin Chaudhuri Jan 25 '13 at 19:48
  • $\begingroup$ I meant i/k above. $\endgroup$ – Arin Chaudhuri Jan 25 '13 at 19:54

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