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Here is my version of the change of variable theorem:

Change of variables: Let $f: [a,b] \to \mathbb{R}$ have an antiderivative and be integrable. Let $g: [c,d] \to \mathbb{R}$ have a derivative in all of its domain with $g'$ integrable and $g([c,d]))\subseteq [a,b]$. Then:

$$\int_{g(c)}^{g(d)}f(x)dx=\int_c^df(g(t))\cdot g'(t)dt$$

I think I understand the proof just fine, just applying the fundamental theorem of calculus two times and checking both sides evaluate to the same thing. However my question is, how do we know $f(g(t))\cdot g'(t)$ is integrable? Since $g'(t)$ is integrable, it is enough to prove that $f(g(t))$ is integrable too. Since $g(t)$ is continuous then it is integrable, however the composition of two integrable functions is not, in general, integrable (proof). So the condition of having a derivative must play some role I believe. That or my book is wrong and the integrability of $f(g(t))$ must be added as a supposition.

Fernando Chu
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    Are you really able to prove this with the FTC only assuming that $f$ is Riemann integrable and $g'$ is Riemann integrable? – RRL Apr 27 '21 at 00:25

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The "easy" proof using the fundamental theorem of calculus requires a the assumption that $f$ is continuous and $g$ is continuously differentiable. Defining $F(t) = \int_{g(c)}^{g(t)}f(x) \, dx$, we have $F'(t) = f(g(t)) g'(t)$ since $f$ is continuous, and

$$\int_{g(c)}^{g(d)}f(x) \, dx = F(d) =F(d)-F(c)= \int_{c}^d F'(t) \, dt = \int_c^d f(g(t))g'(t) \, dt$$

In this case it is also obvious that $f\circ g$ is Riemann integrable.

If we drop the condition that $f$ is continuous and assume only that it is Riemann integrable and assume that $g$ is both continuously differentiable and monotone, another relatively easy proof is facilitated. A proof of the change of variables theorem using Riemann sums is given here. There is no need to prove directly that $f\circ g$ is Riemann integrable.

If $f$ is continuous and $g$ is Riemann integrable, then so is $f\circ g$. What you seem to be concerned with is the case where $f$ is Riemann integrable and $g$ is continuous. Then $f\circ g$ may fail to be Riemann integrable (under even stronger conditions than Riemann integrability of both $f$ and $g$ that you mention).

However, if in addition to $g \in C^1([c,d]$), the derivative $g'$ is non-zero. then we 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, $D_{f \circ g} = g^{-1}(D_f)$.

RRL
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    Kestelman proves a more general theorem in Mathematical Gazette, 45(1961), pages 17-23: Assume $h$ Riemann integrable on $[c, d] $ and $a\in[c, d], g(x) =\int_a^x h(t), dt$. If $f$ is Riemann integrable on $g([c, d]) $ then $\int_c^d f(g(t)) h(t) , dt$ exists and equals $\int_{g(c)} ^{g(d)} f(x) , dx$. In the proof Kestelman mentions that it may be possible that $f(g(t)) $ is not Riemann integrable, but the product $f(g(t)) h(t) $ is Riemann integrable. That I found really surprising and interesting. – Paramanand Singh Apr 27 '21 at 03:43
  • @ParamanandSingh: Thanks. That adds a lot here. I knew of the theorem but overlooked that fact about integrability of $f \circ g$ which is indeed surprising. – RRL Apr 27 '21 at 03:59
  • Yeah using that theorem is an overkill here. I was merely trying to draw attention to the integrability of $f\circ g$. There are a lot of weird cases if one analyzes Riemann integral. That's probably why they developed the Lebesgue integral. – Paramanand Singh Apr 27 '21 at 06:05