Here is theorem 7.7 from Tom Apostol, Mathematical Analysis:
Let $f\in \mathfrak R (\alpha)$ on $[a,b]$ and let $g$ be a strictly monotonic continuous function defined on an interval $S$ having enpoints $c$ and $d$. Assume that $a=g(c)$,$b=g(d)$. Let $h$ and $\beta$ be the composite functions defined as follows: $$h(x)=f[g(c)]$$ $$ \beta (x)=\alpha [g(x)]$$, if $x\in S$.
Then $h\in \mathfrak R (\beta )$ on $S$ and we have $\int_a ^b f d\alpha = \int_c ^d h d\beta $. That is, $$ \int_{g(c)} ^{g(d)} f(t) d\alpha (t) = \int_c ^d f[g(x)] d( \alpha [g(x)]) $$
And the proof:
The question:
Do we really need to make the assumption that $g$ is continuous ?
Apostol says $g$ must be continuous, but I nowhere see he uses it in the proof.
In Rudin's Principles of Mathematical Analysis, theorem $6.19$ on change of variable, it is the same: he states $g$ has to be continuous in the hypothesis, but I don't see the hypothesis used in the proof.
T.I.A.
Its a theorem for an interval of monotonicity only, that is easy to prove.
In the general case the function has to be absolutely continuous, that is, it is the derivative of an integral.
And if it has (discrete points of ) discontinuouities, by additivity, its the sum over intervals of continuity.
Finally, the integral of a non strict monotone function has to be partitioned as a sum over monotone increasing, constant and monotone decreasing intervals.
All this is immediately transparent, if integration over $dy$ is used: Lebesgue integral $\int y \ \mathbb{vol}( f^{-1}(dy) ))$ where the preimage of $dy$ automatically decomposes in coherent subsets of the real $x$-line. And all limits are trivial, including the fact that the integral can be formally extended to all of $\mathbb R$.