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This theorem is from Strichartz' The Way of Analysis:

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Where do we need to use the hypothesis that $g$ is strictly increasing? I think all we need is $g(a)<g(b)$ and $g(x)\in [g(a),g(b)]$ for all $x\in [a,b]$. Am I right?

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  • $\begingroup$ You are right. We don't need any hypotheses about monotone nature of $g$. Just the continuity of $f$ and $g'$ is sufficient. Also no specific order relation between $g(a), g(b) $ is required. $\endgroup$ – Paramanand Singh Sep 8 '17 at 19:59
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You're not going to like this answer...

because it has almost nothing to do with math. And it's certainly not important for the proof.

Requiring that $g$ is increasing on $[a,b]$ allows sentence two to be written

Then for any continuous function $f$ on $[g(a),g(b)]$, we have...

rather than needing to be written

Then for any continuous function $f$ on $[\min \left( g(a),g(b)\right), \max \left( g(a),g(b)\right)]$, we have...

It makes the presentation simpler/cleaner. That's all.

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    $\begingroup$ It is much better to say that $f$ is continuous on $g([a, b]) $ (the range of $g$). $\endgroup$ – Paramanand Singh Sep 9 '17 at 6:42
  • $\begingroup$ Sorry, I'm not following: are you saying the original author should have said that, or I should be saying that? I.e. can you make clear what improvement you're suggesting I make to this post? $\endgroup$ – nitsua60 Sep 9 '17 at 13:32
  • $\begingroup$ The original author should have mentioned that $f$ is continuous on $g([a, b]) $ instead of what is actually written in the book and I think this would look better / concise than using the $\min, \max$ functions as suggested in your answer. $\endgroup$ – Paramanand Singh Sep 9 '17 at 14:47

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