As I have seen in analysis books, the change-of-variables theorem for Riemann integrals gives
$$ \int_{g(a)}^{g(b)} f(x) \,dx = \int_a^b f(g(x))g'(x)\,dx$$
Conditions for $g:[a,b] \to \mathbb{R}$ and $f: g([a,b]) \to \mathbb{R}$ are usually that $f$ is continuous and $g$ is continuously differentiable.
The proof defines $F(x) = \int_{g(a)}^{g(x)} f(t)\,dt$ and since $f$ is continuous we use FTC to get $F'(x) =f(g(x))g'(x)$. Also $F(a) = 0$ so that
$$\int_{g(a)}^{g(b)} f(x)\,dx = F(b) = F(b)-F(a) = \int_a^bF'(x)\,dx = \int_a^bf(g(x))g'(x)\,dx$$
Question: I also read somewhere that if $g$ is a monotonic function we can drop the assumption that $f$ is continuous and only assume it is integrable. How is this proved?
The MVT makes it possible to write $$\left|\int_{g(a)}^{g(b)} f - \sum_{i=1}^n f(g(\xi_i)) g'(\xi_i) (x_i-x_{i-1})\right| \le $$$$ \left|\int_{g(a)}^{g(b)} f - \sum_{i=1}^n f(g(\xi_i))(g(x_i)-g(x_{i-1})) \right| + \left|\sum_{i=1}^n f(g(\xi_i))(g'(\eta_i)-g'(\xi_i))(x_i-x_{i-1})\right|$$ Assuming what you say in your question, it is only about to play the usual $\varepsilon$-$\delta$ game, remembering $f$ is bounded and $g$ and $g'$ are uniformly continuous.
Of course every tagged partition of $[a,b]$ induces a tagged partition of $[g(a),g(b)]$ through $g$ .