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Let $f$ be a Riemann integrable function on $[a,b]$ and $g: [\alpha, \beta] \to [a, b]$ be a continuously differentiable function. Then how can I prove that $f \circ g$ is Riemann integrable? I think this is an easy task, but I can't get it. Maybe uniform continuity of $f$(+uniform continuity of $f'$) seems to be crucial, but I can't proceed any more.

There are no more assumptions about $f$ other than stated. According to the textbook containing this(Krantz, Real Analysis and Foundations, 4th ed, sec 7.3. It is an elementary real analysis textbook), this is an 'easier result' than proving that 'composition $\phi \circ f$ of continuous function $\phi$ on compact interval and Riemann integral function $f$ on $[a,b]$ is Riemann integrable'. I would appreciate answers not using measure theory if possible.

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  • $\begingroup$ Checking an earlier edition, the author posed the problem with the assumption only that $g$ is continuous. The statement that the composition $f \circ g$ must be Riemann integrable is clearly false with the usual counterexample using a fat Cantor set. An attempt to correct the problem in a later edition added continuity of $g’$ but that is not enough and the counterexample is similar. As I showed an additional assumption that ensures that $g$ is homeomorphic makes the statement true. $\endgroup$
    – RRL
    Dec 5, 2020 at 15:10
  • $\begingroup$ I have seen many problems in textbooks that are incorrectly stated or not solvable with the material covered prior to the appearance of the problem. $\endgroup$
    – RRL
    Dec 5, 2020 at 15:14
  • $\begingroup$ @RRL I don't think a homeomorphism is enough; such a map can take a set of positive measure to a set of zero measure. $\endgroup$
    – zhw.
    Dec 5, 2020 at 19:16
  • $\begingroup$ @zhw: Yes - more like absolutely continuous which is the case when the derivative of $g^{-1}$ is continuous and bound on a compact interval. $\endgroup$
    – RRL
    Dec 5, 2020 at 19:40
  • $\begingroup$ If $g^{-1}$ is absolutely continuous then it maps measure zero sets to measure zero sets. $\endgroup$
    – RRL
    Dec 5, 2020 at 19:47

2 Answers 2

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This is false. Let $\alpha=a=0, \beta = b = 1.$ Define $f=\chi_{(0,1]}.$ Then $f$ is Riemann integrable on $[0,1].$

To define $g,$ we use the following fact: Given a closed subset $E\subset \mathbb R,$ there is a $C^\infty$ function $g:\mathbb R\to [0,1]$ such that $g=0$ on $E$ and $g>0$ everywhere else.

With that in mind, choose a nowhere dense compact set $K\subset [0,1],$ with $m(K)>0.$ Use $E=K$ in the last paragraph to obtain $g\in C^\infty$ with the properties above.

Claim: $f\circ g$ is discontinuous at each $x\in K.$

Proof: Let $x\in K.$ Because $K$ is nowhere dense, there exists a sequence $x_n \in [0,1]\setminus K$ with $x_n\to x.$ We then have $g(x_n)>0$ for all $n.$ Hence $f\circ g(x_n) = 1$ for all $n.$ But $x\in K$ implies $g(x)=0,$ which gives $f\circ g(x)=0.$ This gives the claim.

Now if $f\circ g$ were Riemann integrable on $[0,1],$ then its set of discontinuities would have measure $0.$ But the claim implies $f\circ g$ is discontinuous on a set of positive measure, namely $K.$ This is a contradiction, proving $f\circ g$ is not Riemann integrable

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  • $\begingroup$ Though I don't know about measure theory, I can understand what you are saying. But I have a question in the second paragraph. I know that Given a closed subset $E \subset R$, there is a continuous function $f$ that vanishes precisely on $E$(that is related to the Urysohn Lemma). But how can I assure that it is $C^\infty$? $\endgroup$
    – Sphere
    Dec 4, 2020 at 11:59
  • $\begingroup$ @Sphere The complement of a closed set is open, hence is the countable union of disjoint open intervals. On $(a,b)$ the function $e^{-1/[(x-a)(b-x)]}$ leads to a $C^\infty$ function positive on $(a,b)$ and $0$ elsewhere. $\endgroup$
    – zhw.
    Dec 5, 2020 at 18:59
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This is true if in addition to $g \in C^1([\alpha,\beta]$), the derivative $g'$ is non-zero. We then have an inverse function $g^{-1}$ which is continuously differentiable, mapping sets of measure zero into sets of measure zero.

Since $f$ is Riemann integrable it is bounded and the set of discontinuity points $D_f$ has measure zero. Thus, $f \circ g$ is Riemann integrable since it is bounded with discontinuity points in a set of measure zero, $D_{f \circ g} = g^{-1}(D_f)$.

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  • $\begingroup$ Thanks, but the original question doesn't assume that $g'$ is nonzero. Also, I haven't studied about (Lebesgue) measure theory. $\endgroup$
    – Sphere
    Dec 1, 2020 at 4:48
  • $\begingroup$ Measure zero is a pretty basic concept — fundamental to characterizing Riemann integrable functions. This is not venturing far into “measure theory”. Why not understand it? What is the problem source? Is it homework and you can’t use what does not appear in the problem’s chapter? The history of problems related to this on this site suggests it is not trivial. For example the counterexample when $g$ is only continuous brings up Cantor-like sets. $\endgroup$
    – RRL
    Dec 1, 2020 at 11:46
  • $\begingroup$ I haven't learned about measure at all yet, but I will try. How about the case where $g'$ has a zero point? $\endgroup$
    – Sphere
    Dec 1, 2020 at 13:16
  • $\begingroup$ It is crucial that $g’ \neq 0$. $\endgroup$
    – RRL
    Dec 5, 2020 at 15:23
  • $\begingroup$ Certainly $g'$ could be zero at one point and still preserve Riemann integrability. $\endgroup$
    – zhw.
    Dec 18, 2020 at 18:48

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