Let $g$ be a Riemann integrable function on $[a,b]$, and assume $f$ is a continuous function defined on $g(x)$, for all $x\in[a,b]$. Prove that $f(g(x))$ is Riemann integrable for all $x\in[a,b]$.

What I have so far: I know that if $f(g)$ is continuous then certainly Riemann, but this requires $g$ to be continuous. So consider the definition of Riemann integrable: for all $\epsilon>0$ there exists a partition $P$ such that the difference of upper sum and lower sum is bounded by this $\epsilon$. Since $g$ is Riemann integrable, there exists a partition such that $U(P,g)-L(P,g)<\epsilon$. Since $f$ is continuous, there exists $\delta>0$ such that the difference of upper sum and lower sum over any interval of length $\delta$ is bounded by $\epsilon$. I'm having trouble of combining these two together...


2 Answers 2


Consider a partition $a = x_0 < x_1 < \ldots < x_n = b$ and for a subinterval $I_j = [x_{j-1},x_j]$ define

$$D_{j}(g) = \sup_{x \in I_j}g(x) - \inf_{x \in I_j}g(x) = \sup_{x,y \in I_j}|g(x) - g(y)| \\ D_{j}(f \circ g ) = \sup_{x \in I_j}f(g(x)) - \inf_{x \in I_j}f(g(x)) = \sup_{x,y \in I_j}|f(g(x)) - f(g(y))|$$

Since $f$ is continuous it is uniformly continuous and bounded (by extension if necessary) on a closed interval $[c,d]$ such that $g([a,b]) \subset [c,d].$

Hence, $|f(x)| \leqslant M$ for $x \in [c,d]$ and for every $\epsilon >0$ there exists $\delta > 0$ such that if $|x_1 - x_2| < \delta$ then $|f(x_1) - f(x_2)| < \epsilon/(2(b-a))$ .

Since $g$ is integrable, if the partition norm $\|P\|$ is sufficiently small we have

$$U(P,g) - L(P,g) = \sum_{j=1}^n D_j(g) (x_j - x_{j-1}) < \frac{ \delta \epsilon}{4M}.$$

We can split the upper-lower sum difference $U(P,f \circ g) - L(P, f \circ g)$ into two sums as given by

$$\tag{1}U(P,f \circ g) - L(P, f \circ g) = \sum_{D_j(g) \geqslant \delta} D_j(f \circ g)(x_j - x_{j-1}) + \sum_{D_j(g) < \delta} D_j(f \circ g)(x_j - x_{j-1})$$

In the second sum on the RHS of (1) we have $D_j(f \circ g) < \epsilon/(2(b-a)$ since by uniform continuity $D_j(g) < \delta \implies |g(x) - g(y)| < \delta \implies |f(g(x)) - f(g(y))| < \epsilon/(2(b-a)$ for all $x,y \in I_j$.


$$\tag{2}\sum_{D_j(g) < \delta} D_j(f \circ g)(x_j - x_{j-1}) < \frac{\epsilon}{2}$$

Considering the first sum on the RHS of (1), first note that

$$\sum_{D_j(g) \geqslant \delta} (x_j - x_{j-1}) \\ < \delta^{-1}\sum_{D_j(g) \geqslant \delta} D_j(g)(x_j - x_{j-1}) < \delta^{-1} [U(P,g) - L(P,g)] < \delta^{-1} \frac{\delta \epsilon}{4M} = \frac{\epsilon}{4M} .$$


$$\tag{3}\sum_{D_j(g) \geqslant \delta} D_j(f \circ g)(x_j - x_{j-1}) < \sum_{D_j(g) \geqslant \delta} 2M(x_j - x_{j-1}) < \frac{\epsilon}{2}.$$

From (1), (2) and (3) we obtain

$$U(P,f \circ g) - L(P, f \circ g) < \epsilon,$$

and conclude that $f \circ g$ is integrable.

  • $\begingroup$ Thank you for your comment, but why is $f$ continuous on a closed interval $[c,d]$? What do you mean by "by extension if necessary?" $\endgroup$ Commented Sep 22, 2017 at 1:20
  • $\begingroup$ Clearly since $g$ is Riemann integrable it is bounded and the image $g([a,b])$ is a bounded set. You say that $f$ is continuous and nothing more. At a minimum to proceed, this should mean that $f$ is continuous on $g([a,b])$. If that set is closed (hence compact) or $f$ can be taken to be a continuous function on the closure of $g([a,b])$ then we have the what is required to invoke the uniform continuity and boundedness used in the proof without even referring to an interval $[c,d].$ I'm not sure how far you want to delve into the nature of $g([a,b])$ but I'm sure that can be resolved. $\endgroup$
    – RRL
    Commented Sep 22, 2017 at 2:41
  • $\begingroup$ So if we add that $f$ is continuous on the closure, we are done. The point of this answer is to show how this can be proved with Darboux sums and the Riemann criterion for integrability. I personally find that more instructive than simply invoking Lebesgue. $\endgroup$
    – RRL
    Commented Sep 22, 2017 at 2:45
  • 1
    $\begingroup$ +1 for doing it the hard but more elementary way. The other approach via sets of measure zero appears to be a one liner, but it depends on a highly non-trivial theorem that a function is Riemann integrable iff and only its discontinuities form a set of measure zero. $\endgroup$
    – Paramanand Singh
    Commented Sep 22, 2017 at 16:43
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    $\begingroup$ @TranKhanh: Yes. The Riemann criterion for integrability is that for any $\epsilon > 0$ there exists a partition $P_\epsilon$ such that the difference between upper and lower Darboux sums satisfies $U(P_\epsilon,g) - L(P_\epsilon,g) < \epsilon$. We only need to show the existence of one such partition given $\epsilon$ and the parameters used in the construction of that partition don't matter. $\endgroup$
    – RRL
    Commented Jan 31 at 18:23

EDIT: I wrote the solution below under the implicit assumption that $f$ is defined and continuous on $\mathbb{R}$. But this is not clearly stated in the question, and if absolute continuity does not hold the boundness need not hold either, as rightfully pointed out by @KilluaZoldyck

Use Lebesgue criterion for Riemann integrability:

$g$ is Riemann integrable iff $g$ is bounded and the sets of its discontinuities has Lebesgue measure zero.

$f$ is continuous, so $f\circ g$ is bounded. Moreover $\operatorname{disc}(f\circ g)\subseteq \operatorname{disc}(g)$, so the set of discontinuities of $f\circ g$ has Lebesgue measure zero.

So $f\circ g$ is Riemann integrable.

  • 1
    $\begingroup$ '$f$ is continuous so $f \circ g$ is bounded'. How does this statement follow? It is not true in general that continuous fns map bounded set to bounded sets (only true if a fn is uniformly continuous). Can you please explain why that statement is true? $\endgroup$ Commented Dec 8, 2021 at 11:50
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    $\begingroup$ @KilluaZoldyck Good catch. If $g:[0,1]\to(0,1]$ with $g(0)=1$ and $g(x)=x$ for $x\ne 0$ and $f:(0,1]\to\mathbb{R}$, $f(x)=1/x$, then $g$ is obviously Riemann integrable and $f$ is continuous, but $f\circ g$ is not bounded, therefore not Riemann integrable. Definitely uniform continuity is needed. $\endgroup$
    – Momo
    Commented Dec 8, 2021 at 22:54

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