Composition of Riemann-integrable and increasing functions.

Question

This is exercise 7.3.3 from Abbot's Understanding analysis. The section is Integrating functions with discontinuities.

I am struggling with this exercise. I can't either come up with any simple counterexamples or produce a satisfactory proof.

I was able to solve a), and here are some answers to b) that I don't understand well because they use a bit of measure theory: Answer 1, Answer 2. I still don't know whether c) is true or false.

Any hints to a proof, the proof itself, or simple counterexamples are highly appreciated.

The exercise reads:

Let $f$ and $g$ be functions defined on (possibly different) closed intervals, and assume the range of $f$ is contained in the domain of $g$ so that the composition $g \circ f$ is properly defined.

a) Show, by example, that it is not the case that if $f$ and $g$ are integrable, then $g\circ f$ is integrable.

Now decide on the validity of each of the following conjectures, supplying a proof or counterexample as appropriate.

b) If $f$ is increasing and $g$ is integrable, then $g\circ f$ is integrable.

c) If $f$ is integrable and $g$ is increasing, then $g\circ f$ is integrable.

EDIT

I found a solution to b) here, due to Chutong Wu. The solution seems to contradict the counterexamples from above, as it provides a proof of the statement. I reproduce the proof in full:

Fix $\epsilon > 0$. Because $g$ is integrable, we can find a partition $P_{g}=\{x_{0}<\cdots <x_{n}\}$ of $Rng(f)\subseteq Dom(g)$ such that $U(g\lvert_{Rng(f)}, P_{g})-L(g\lvert_{Rng(f)}, P_{g})<\epsilon$.

Because $f$ is increasing, it is one-to-one so $f^{-1}:Rng(f)\rightarrow Dom(f)$ is well-defined and also increasing. This means the set $P=\{f^{-1}(x_{0}),\cdots,f^{-1}(x_{n})\}$ is also a partition of $Dom(f)$.

We then have:

$ \begin{align} U(g\circ f, P)-L(g\circ f, P) &=\sum_{k=1}^{n}( \underbrace{\sup_{[x_{k-1},x_{k}]} g\circ f}_{=(g\circ f)(f^{-1}(x_{k}))=g(x_{k})} - \underbrace{\inf_{[x_{k-1},x_{k}]} g\circ f}_{=(g\circ f)(f^{-1}(x_{k}))=g(x_{k-1})}) [x_{k-1},x_{k}] \\ &= U(g\lvert_{Rng(f)}, P_{g})-L(g\lvert_{Rng(f)}, P_{g})<\epsilon \end{align} $

Answer 1

(c) Let $\phi:\mathbb{N}\rightarrow(0,\infty)$ be a monotone nonincreasing sequence such that $\phi(n)\xrightarrow{n\rightarrow\infty}0$. Define $f:[0,1]\rightarrow\mathbb{R}$ as follows $$f(x)=\left\{\begin{array}{lcr}1 &\text{if} & x=0\\ 0 &\text{if} & x\in[0,1]\setminus\mathbb{Q}\\ \phi(n) &\text{if}& x=\frac{m}{n},\quad (m,n)=1 \end{array} \right. $$

The case $\phi(x)=\frac{1}{x}$ yields what is know as Thomae's function. It can be seen that $f$ is continuous at irrational points. Thus $f$ is $R$-integable over $[0,1]$ (by Lebesgue's criteria).

Consider the function $g(x)=\mathbb{1}_{(0,\infty)}(x)$. This function is monotone nondecreasing, and $$h(x)=g(f(x))=\mathbb{1}_{[0,1]\cap\mathbb{Q}}(x)$$ which is not $R$-integrable.

By considering $G(x)=x+g(x)$, we obtain an strictly monotone increasing function such that $G\circ f$ is not $R$-integrble.

---

Just for completion, here is a proof that the Thomae-like function $f$ defined above is indeed continuous at every $x\in[0,1]\setminus\mathbb{Q}$. Fix $x_0\in [0,1]\setminus\mathbb{Q}$ and $\varepsilon>0$. Let $r\in\mathbb{R}$ such that $\phi(r) <\varepsilon$. For each $j\in\{1,\ldots r\}$ let $k_j=\lfloor jx_0\rfloor$. Since $x_0$ is irrational $$k_j<jx_0<k_j+1$$ Let $\delta:=\min_{1\leq j\leq r}\left\{\big|x_0-\frac{k_j+1}{j}\big|,\big|x_0-\frac{k_j}{j}\big|\right\}$. Suppose $\operatorname{g.c.d}(p,q)=1$ and $|x-p/q|<\delta$. We claim that $q>r$, otherwise $q\leq r$ and so $p\leq k_q$ or $p\geq k_p+1$. This in turn implies that $$\big|x-\frac{p}{q}\big|\geq\delta$$ which leads to a contradiction. Therefore, if $|x-p/q|<\delta$, $$|f(p/q)-f(x)|=f(p/q)=\phi(q)\leq \phi(r)<\varepsilon$$

Answer 2

This is to address some comments by the OP regarding a proof to (b) that he found in the web. Other members of the community are welcome to add/improve this posting.

Suppose $f:[\alpha,\beta]\rightarrow\mathbb{R}$, $g:[a,b]\rightarrow\mathbb{R}$ are $R$-integrable, that $A:=f([\alpha,\beta])\subset[a,b]$ and that $f$ is monotone nondecreasing (strictly monotone increasing seems to be what the OP assumes, but for the time being the previous assumptions will do).

1. The solution to (b) presented by Chutong Wu is not correct. There are at least to big error.

• (a) Notice that even if $A$ were an interval, say $[c,d]\subset [a,b]$ and $f$ bijective. The error in Wu's argument stems from the fact if $\mathcal{P}=\{c=f(\alpha)=y_0<\ldots<y_n=d=f(\beta)\}$, and $\mathcal{P}'=f^{-1}(\mathcal{P})=\{\alpha=x_0<\ldots x_n=\beta\}$, the Riemann-Darboux sums for $g$ $$\begin{align} U(g;\mathcal{P})&=\sum^n_{j=1}\big(\sup_{y\in[y_{j-1},y_j]}g(y)\big)(y_j-y_{j-1})\\ &=\sum^n_{j=1}\big(\sup_{x\in[x_{j-1},x_j]}g(f(x))\big)(f(x_j)-f(x_{j-1}))\\ L(g;\mathcal{P})&=\sum^n_{j=1}\big(\inf_{y\in[y_{j-1},y_j]}g(y)\big)(y_j-y_{j-1})\\ &=\sum^n_{j=1}\big(\inf_{x\in[x_{j-1},x_j]}g(f(x))\big)(f(x_j)-f(x_{j-1})) \end{align}$$ are not Riemann-Darboux sums for $g\circ f$, that is, they are not of the form $$\begin{align} U(g\circ f;\mathcal{P}')&=\sum^n_{j=1}\big(\sup_{x\in[x_{j-1},x_j]}g(f(x))\big)(x_j-x_{j-1})\\ L(g\circ f;\mathcal{P}')&=\sum^n_{j=1}\big(\inf_{x\in[x_{j-1},x_j]}g(f(x))\big)(x_j-x_{j-1}) \end{align}$$
• (b) The Riemann integral is built on partitioning bounded closed intervals, not arbitrary sets. That is, the $R$ integral of $g\mathbb{1}_A$ is build by partinoning $[a,b]$ (or any other subinterval that contains $A$), and not $A$ itself.

2. The set $A$ may not be nice for Riemann integration, even when $g$ is $R$-integrable over $[a,b]$. Take for example $g(x)=x$ and $A=\mathbb{Q}\cap[0,1]$. The function $g\mathbb{1}_A$ is not $R$-integrable in $[0,1]$.
3. The assumption that $f$ is $R$-integrable and monotone increasing over say, $[\alpha,\beta]$, implies that $f$ admits at most a countable number of discontinuities (all of which are jump discontinuities). Still, $f([\alpha,\beta])$ may not be interval or even the countable union of and ordered collection of closed intervals and points; in fact, it could be a rather nasty set (still nice in the sense of measure theory). Take for example Cantor's function (the devil's staircase) $F$. Define $Q:[0,1]\rightarrow[0,1]$ as $Q(y)=\inf\{x\in [0,1]: F(x)=F(y)\}$. It can bee seen that $Q$ is strictly monotone increasing (actually strictly increasing) and that $Q([0,1])$ is the Cantor set. $Q$ is continuous on $[0,1]\setminus D$, where $D=\{m2^{-n}: n\in\mathbb{Z}_+, m\in\mathbb{Z}_+\}\cap[0,1]$ is the set of dyadic numbers.

---

4. The counterexample to (b) presented here is correct. The crux of the matter is to check that indeed there is a continuous strictly monotone increasing function $f:[0,1]\rightarrow[0,1]$ that maps a fat Cantor set $S$ onto the 1/3-Cantor set $C$. For then, $h(x)=\mathbb{1}_{C}(f(x))$ is discontinuous at every point $x\in S$. Since $S$ is not of (Lebesgue) measure $0$, then $h$ is not R-integrable and yet $\mathbb{1}_C$ and $f$ are $R$-integrable in $[0,1]$.