# Integrability of composite of Riemann integrable functions given condition

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 $$g$$.

• You can try to use the criterion for Riemann integrability given by Riemann. – Paramanand Singh Jan 15 at 7:24
• My comment about monotonicity was incorrect, it is easy to create a counterexample. – copper.hat Jan 16 at 14:23
• @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. – Paramanand Singh Jan 17 at 10:40
• @ParamanandSingh: $g(x) = \begin{cases} {2 \over 3} x, & x \in [0,{1 \over 2}) \\ 1-{2 \over 3}(x-{1 \over 2}),& x \in [{1 \over 2},1]\end{cases}$. – copper.hat Jan 17 at 16:11
• See this thread math.stackexchange.com/q/2463714/72031 which also assumes continuity of $g$. – Paramanand Singh Jan 18 at 1:57

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

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

$$\frac{1}{\alpha}\sum_i(M_i-m_i)(y_{i+1}-y_i).$$

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 $$\epsilon.$$

• Why is $g$ strictly monotone? $g$ is known to be one-one but that does not guarantee monotone nature unless it is also continuous. – Paramanand Singh Jan 16 at 2:49
• @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. – RRL Jan 16 at 20:32
• @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. – Paramanand Singh Jan 17 at 1:20
• @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. – Paramanand Singh Jan 17 at 1:31
• @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. – copper.hat Jan 17 at 16:22