This is a question from a past qualifying exam, which I am studying for. The question has been asked before here, and has an answer, but the answer uses Lebesgue's criterion for Riemann integrability, which is disallowed on the exam. Is there a more elementary way to solve this question?

Let $f: [0,1] \to \mathbb{R}$ and $g: [0,1] \to [0,1]$ be two Riemann integrable functions. Assume that $|g(x) - g(y)| \geq \alpha |x-y|$ for any $x,y \in [0,1]$ and some fixed $\alpha \in (0,1)$. Show that $f \circ g$ is Riemann integrable.

Some thoughts have been bounding the intervals in which $f$ has a large oscillation by its integrability, and trying use the condition on $g$ to control the growth of these interval lengths. However, I am unsure how to apply the Riemann integrability of $g$.

  • 3
    $\begingroup$ You can try to use the criterion for Riemann integrability given by Riemann. $\endgroup$
    – Paramanand Singh
    Jan 15, 2020 at 7:24
  • 1
    $\begingroup$ My comment about monotonicity was incorrect, it is easy to create a counterexample. $\endgroup$
    – copper.hat
    Jan 16, 2020 at 14:23
  • $\begingroup$ @copper.hat: Can you provide the counter-example? There are some arguments (see comments to the answer by mate at leta) which indicate that $g$ should be monotone. $\endgroup$
    – Paramanand Singh
    Jan 17, 2020 at 10:40
  • 1
    $\begingroup$ @ParamanandSingh: $g(x) = \begin{cases} {2 \over 3} x, & x \in [0,{1 \over 2}) \\ 1-{2 \over 3}(x-{1 \over 2}),& x \in [{1 \over 2},1]\end{cases}$. $\endgroup$
    – copper.hat
    Jan 17, 2020 at 16:11
  • $\begingroup$ See this thread math.stackexchange.com/q/2463714/72031 which also assumes continuity of $g$. $\endgroup$
    – Paramanand Singh
    Jan 18, 2020 at 1:57

1 Answer 1


I have not worked through all of the details, but here is a sketch of an idea, too long for a comment. I will put question mark at the part I have not thought through yet.

Set $I:=[0,1]$ and choose an integer $k$ such that $\frac{1}{k}<\frac{\epsilon}{2}.$ The set $D_k=\{x\in I:\text{osc}_x\ f\ge1/k\}$ has measure zero so it has a countable covering by open sets $J_j = (a_j, b_j)$ whose total length is less than $\frac{\epsilon}{2}.$ Now, for each $x\in I\setminus D_k$ there is an open interval $x\in I_x\subseteq I\setminus D_k$ such that $\sup\ f-\inf\ f<1/k$ on $I_x$ (because $\text{osc}_x\ f<1/k$). Then, the $J_j$ and $I_x$ form an open cover of $I$. Let $\lambda$ be the Lebesgue number of this cover and take any partition $Q=\{y_i\}$ of $I$ such that $|Q|<\lambda$ and

$[y_i,y_{i+1}]\subseteq \text{im}\ g$. ???

Let $M_i,m_i$ be the maxima, resp. minima of $f$ on $[y_i,y_{i+1}].$

Then let $x_i=g^{-1}(y_i)$. Since $g$ is injective, the $x_i$ form a partition $P$ of $I$ and

$U(f\circ g,P)-L(f\circ g,P)=\sum_i(M_i-m_i)|g^{-1}(y_{i+1})-g^{-1}(y_i)|\le$


By construction, $[y_i,y_{i+1}]$ is either in one of the $J_j$ or one of the $I_x.$ Now split this sum up into those subintervals of $P$ that lie in one of the $J_j$ and those that lie in one of the $I_x$. The set-up in first paragraph shows that the sum is less than $\epsilon.$

  • $\begingroup$ Why is $g$ strictly monotone? $g$ is known to be one-one but that does not guarantee monotone nature unless it is also continuous. $\endgroup$
    – Paramanand Singh
    Jan 16, 2020 at 2:49
  • $\begingroup$ @ParamanandSingh: I believe that we have a case where $g^{-1}$ is injective and Lipschitz continuous which implies $g^{-1}$ is strictly monotone. Now $x < y, \,g(x) \geqslant g(y) \implies g^{-1}(g(x)) \geqslant g^{-1}(g(y)) \implies x \geqslant y$ a contradiction, so $g$ is also strictly monotone. $\endgroup$
    – RRL
    Jan 16, 2020 at 20:32
  • $\begingroup$ @RRL: thanks man! I never looked at $g^{-1}$ although it was mentioned in the thread linked in question. The fact that $g$ is monotone makes the problem far simpler. $\endgroup$
    – Paramanand Singh
    Jan 17, 2020 at 1:20
  • $\begingroup$ @RRL: there is another issue which can also be tackled. The range of $g$ should also be an interval and then only we can use the theorem "one one and continuous implies monotone". In fact it should be possible to prove that $g([0,1])=[0,1]$. But I am not sure. $\endgroup$
    – Paramanand Singh
    Jan 17, 2020 at 1:31
  • 2
    $\begingroup$ @RRL: It must be true that $g^{-1}$ is Lipschitz continuous on $g([0,1])$, but it does not follow that $g$ is monotone as the range of $g$ is not necessarily connected. $\endgroup$
    – copper.hat
    Jan 17, 2020 at 16:22

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