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I was working on some problems on Integration part of the book Problems in Mathematical Analysis by Kaczor and Nowak. And it seems that I don't get this one solution.enter image description here

The sentences underlined with red are unclear to me.If there are $k_N$ rationals then how is it possible to be $2k_N$ subintervals containing at least one of those rationals.Any help would be appreciated. Thank you!

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    $\begingroup$ If one of those points is equal to an $x_i$ it will be counted in $[x_{i-1},x_i]$ as well as $[x_i,x_{i+1}]$. $\endgroup$ Oct 17 '18 at 8:10
  • $\begingroup$ You're right .Thank you very much, and what about the second sentence? $\endgroup$ Oct 17 '18 at 8:19
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Notice that $f$ is nonnegative and in fact by density of real number, $0$ is attained in any interval of positive length.

For those intervals that do not contain of those rational value of which we have $q \le N$. We have $q > N$ and hence $\frac1q < \frac1N$. That is $M_j < \frac1N$ and $m_j = 0$. Hence $$M_j - m_j < \frac1N.$$

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