I am reading a calculus textbook and I am having trouble reading a proof that shows that all monotone functions $f: [a, b] \rightarrow \mathbb{R}$ are integrable. This proof is not complete, specifically the textbook wants me to complete the proof as an exercise:

Without loss of generality let $f$ be monotone increasing. We'll divide the interval $[f(a), f(b)]$ up into $n$ equidistant subintervals with length $$h = \frac{f(b) - f(a)}{n}$$ and $y_k := f(a) + kh$ for $k = 0, 1, \dots, n$. Furthermore, let $a_0 := a$ and $$a_k := \sup\{x \in [a,b] : f(x) \leq y_k\}$$ for $k = 1, \dots, n$. Because $f$ is monotone, it follows that $$a = a_0 \leq a_1 \leq \dots \leq a_n = b$$ Furthermore, we have (EXERCISE) $$x \in (a_{k-1}, a_k) \Rightarrow f(x) \in ( y_{k-1}, y_k]$$


I am really unsure if what I did is correct:

If $x < a_k$, then $f(x) \leq f(a_k) \leq y_k$, because $f$ is monotone increasing. So $f(x) \leq y_k$.

If $x > a_{k-1}$, then there can be no such $x$ with $f(x) \leq y_{k-1}$ because that would contradict the supremum property of $a_{k-1}$. So we have $y_{k-1} < f(x)$.

So $$x \in (a_{k-1}, a_k) \Rightarrow f(x) \in ( y_{k-1}, y_k]$$

  • $\begingroup$ i don't see how $f(x) \le y_{k-1}$ contradicts the supremum property of $a_{k-1}$. in fact the proof is much easier than you might think. $\endgroup$ – tortue Aug 6 '18 at 9:28
  • $\begingroup$ @pointguard0 Because of the definition of $a_{k-1}$, it is the "largest" $x$ for which we have $f(x) \leq y_{k-1}$. Any $x$ greater than $a_{k-1}$ does no longer suffice the inequality $f(x) \leq y_{k-1}$. Or am I missing something? $\endgroup$ – user295213 Aug 8 '18 at 13:46
  • $\begingroup$ and hence, $f(x) > y_{k-1}$ for all $x > a_{k-1}$. this is as simple as that $\endgroup$ – tortue Aug 8 '18 at 14:06
  • $\begingroup$ @pointguard0 Yes, that's exactly what I wrote in my original post. $\endgroup$ – user295213 Aug 8 '18 at 14:08
  • $\begingroup$ okay then, sorry, if i misunderstood. upvoted :) $\endgroup$ – tortue Aug 8 '18 at 14:09

You already wrote

If $x < a_k$, then $f(x) \leq f(a_k)$

Now, in the sentence above, make the following replacements:

  • Replace $x$ with $a_{k-1}$
  • Replace $a_k$ with $x$

What inequality do you get?

  • $\begingroup$ $f(a_{k-1}) \leq f(x)$, now what? $\endgroup$ – user295213 Aug 8 '18 at 13:39

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