Here is Theorem 7.35, in Chapter 7, in the book Mathematical Analysis - A Modern Approach to Advanced Calculus by Tom M. Apostol, 2nd edition:
Assume $f \in R$ on $[a, b]$. Let $\alpha$ be a function which is continuous on $[a, b]$ and whose derivative $\alpha^\prime$ is Riemann integrable on $[a, b]$. Then the following integrals exist and are equal:$$\int_a^b f(x) d \alpha(x) \ = \ \int_a^b f(x) \alpha^\prime(x) dx. $$
And, here is Apostol's proof:
By the second fundamental theorem we have, for each $x$ in $[a, b]$,$$\alpha(x) - \alpha(a) = \int_a^x \alpha^\prime (t) dt.$$Taking $g = \alpha^\prime$ in Theorem 7.33 we obtain Theorem 7.35.
Here is the second fundamental theorem of integral calculus (i.e. Theorem 7.34) in Apostol:
Assume that $f \in R$ on $[a, b]$. Let $g$ be a function defined on $[a, b]$ such that the derivative $g^\prime$ exists in $(a, b$ and has the value$$g^\prime(x) = f(x) \qquad \mbox{ for every } x \mbox{ in } (a, b). $$At the endpoints assume that $g(a+)$ and $g(b-)$ exist and satisfy$$g(a) - g(a+) = g(b) - g(b-). $$Then we have$$\int_a^b f(x) dx = \int_a^b g^\prime(x) dx = g(b) - g(a). $$
Finally, here is Theorem 7.33 in Apostol:
If $f \in R$ and $g \in R$ on $[a, b]$, let$$F(x) = \int_a^x f(t) dt, \qquad G(x) = \int_a^x g(t) dt \qquad \mbox{ if } x \in [a, b]. $$Then $F$ and $G$ are continuous functions of bounded variation on $[a, b]$. Also, $f \in R(G)$ and $g \in R(F)$ on $[a, b]$, and we have$$\int_a^b f(x) g(x) dx = \int_a^b f(x) dG(x) = \int_a^b g(x) dF(x). $$
Now here is my reading of Apostol's proof of Theorem 7.35:
We have$$\alpha(x) = \alpha(a) + \int_a^x \alpha^\prime(t) dt = \alpha(a) + G(x) \qquad \mbox{ for every } x \in [a, b],$$where$$G(x) = \int_a^x \alpha^\prime(t) dt. $$As the integrals $\int_a^b f(x) dx$ and $\int_a^b \alpha^\prime(x) dx$ both exist, so by Theorem 7.33 $\int_a^b f(x) dG(x)$ also exists and therefore $\int_a^b f(x) \alpha^\prime(x) dx$ also exists. Then $\int_a^b f(x) d \alpha(x)$ exists because we have\begin{align}\int_a^b f(x) d\alpha(x) &= \int_a^b f(x) d \big( \alpha(a) + G(x) \big) \\&= \int_a^b f(x) d \alpha(a) + \int_a^b f(x) d G(x) \qquad [\mbox{ by Theorem 7.3 in Apostol }] \\&= 0 + \int_a^b f(x) d G(x) \qquad [\mbox{ because the integrator $\alpha(a)$ in the first integral is a constant }] \\&= \int_a^b f(x) d G(x) \\&= \int_a^b f(x) g(x) dx \qquad [\mbox{ by Theorem 7.33 in Apostol with $g(x) = \alpha^\prime(x)$ }] \\&= \int_a^b f(x) \alpha^\prime(x) dx,\end{align}as required.
Is each and every step in my reading (or rather rendering) of Apostol's proof of Theorem 7.35 correct and accurate as well as clear enough? If so, then what is the necessity of explicitly requiring the function $\alpha$ to be continuoous on $[a, b]$? Exactly where in the proof of this theorem do we need to impose this condition and cannot do without this condition?
Or, is there any problem of accuracy or clarity in any step in my elaborated proof?
PS:
I think this theorem can be more succinctly stated as follows:
Assume $f \in R$ on $[a, b]$, and let $\alpha$ be a function differentiable on $[a, b]$ such that the derivative $\alpha^\prime \in [a, b]$. Then the following two integrals exist and we have$$\int_a^b f(x) d\alpha(x) = \int_a^b f(x)\alpha^\prime(x) dx.$$
Am I right?
And, the proof should go as follows:
As $f, \alpha^\prime \in R$ on $[a, ]$, so we must also have $f \alpha^\prime \in R$ on $[a, b]$, by Theorem 7.33 in Apostol. Let us put $g = \alpha^\prime$. Then the function $G$ of Theorem 7.33 is given by$$G(x) = \int_a^x g(t) dt = \int_a^x \alpha^\prime (t) dt = \alpha(x) - \alpha(a).$$Moreover, we also have $f \in R(G)$ also by Theorem 7.33, and by Theorem 7.3 in Apostol we have$$\int_a^b f(x) dG(x) = \int_a^b f(x) d \big( \alpha(x) - \alpha(a) \big) = \int_a^b f(x) d \alpha(x) - \int_a^b f(x) d\alpha(a) = \int_a^b f(x) d \alpha(x) - 0 = \int_a^b f(x) d\alpha(x),$$so that $f \in R(\alpha)$ on $[a, b]$.
Is this reasoning correct?
Yet again by Theorem 7.33, we get\begin{align}\int_a^b f(x) \alpha^\prime(x) dx &= \int_a^b f(x) g(x) dx \\&= \int_a^b f(x) dG(x) \\&= \int_a^b f(x) d\alpha(x),\end{align}as required.
Is this proof correct and clear enough? If so, isn't it more logically sound than my earlier proof?
