In Chapter 1 of Apostol Calculus, Apostol provides proof of Theorem 1.19, however only shows the proof for the case of k > 0, saying k < 0 can be shown similarly.
I've tried working out the second half on my own, but I'm having some trouble. I'm hoping somebody can provide a hint or pose a question that will help me along. I've copied the case for $k > 0$ below, and then shown what I have done below that. I'll add that I've gone through every page/exercise up until this point in this book, but I have no experience beyond that (i.e. analysis)
Theorem 1.19: Expansion or Contraction of the Interval of Integration
If $f$ is integrable on $[a, b]$, then for every real $k \neq 0$ we have $$\int_{a}^{b}{f(x)dx} = \frac{1}{k}\int_{ka}^{kb}{f(\frac{x}{k})dx}$$
Here is the case for $k > 0$, given by Apostol.
Case: $k > 0$
Assume $k > 0$ and define $g$ on the interval $[ka, kb]$ by the equation $g(x) = f(x/k)$. Let $\underline{I}(g) \text{ and } \overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. We shall prove that
$$ \underline{I}(g) = \overline{I}(g) = k \cdot \int_{a}^{b}{f(x)}dx \label{1.17}\tag{1.17}$$
Let $s$ be any step function below $g$ on $[ka, kb]$. Then the function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ is a step function below f on $[a,b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. Also by the expansion property for integrals of step functions, we have
$$\int_{ka}^{kb}{s(x)dx} = k \cdot \int_{a}^{b}{s(kx)dx} = k \cdot \int_{a}^{b}{s_{1}(x)dx}$$
Therefore we have
$$ \underline{I}(g) = \sup \left\{ \int_{ka}^{kb}{s} \:|\: s \leq g\right\} = \sup \left\{ k \cdot \int_{a}^{b}{s_{1}} \:|\: s_{1} \leq f\right\} = k \cdot \int_{a}^{b}{f(x)dx}$$
Similarly, we find $\overline{I}(g) = k \cdot \int_{a}^{b}{f(x)dx}$ which proves (1.17)
Case: $k < 0$
I'm not entirely sure why the proof needs to be split into two cases, however since it was, my thought was to define $g$ on the interval $[kb, ka]$ since $k < 0$ (I'm imagining g is f, reflected over the y-axis). Then define the step function below $g$ and $f$ along the same lines. Later, in line (1), I make use of the reflection property for when lower limit of integration is greater than the upper limit. However at the end of this, I'm left with an extra negative.
Assume $k < 0$ and $b > a$. Define $g$ on the interval $[kb, ka]$ by the equation $g(x) = f(x/k)$. Let $\underline{I}(g) \text{ and } \overline{I}(g)$ denote the lower and upper integrals of $g$ on $[ka, kb]$. Again we want to show (1.17).
Let $s$ be any step function below $g$ on $[kb, ka]$. Then there is a function $s_{1}$ defined on $[a, b]$ by the equation $s_{1}(x) = s(kx)$ that is a step function below f on $[a, b]$. Moreover, every step function $s_{1}$ below $f$ on $[a, b]$ has this form. By the expansion property for integrals of step functions, we have
$$\int_{kb}^{ka}{s(x)dx} = k \cdot \int_{b}^{a}{s(kx)dx} = k \cdot \int_{b}^{a}{s_{1}(x)dx} = -k \cdot \int_{a}^{b}{s_{1}(x)dx}\label{1}\tag{1}$$
$$ \underline{I}(g) = \sup \left\{ \int_{kb}^{ka}{s} \:|\: s \leq g\right\} = \sup \left\{ -k \cdot \int_{a}^{b}{s_{1}} \:|\: s_{1} \leq f\right\} = -k \cdot \int_{a}^{b}{f(x)dx}\label{2}\tag{2}$$
It's at this point that I'm stuck. Working it backwards, the only way I see to remove the last negative in (2) is to start with $g$ defined on $[ka, kb]$ but then why was the proof split into two cases if they are identical?
One other thought I had was to explore defining, in (2), $$\sup \left\{ -k \cdot \int_{a}^{b}{s_{1}} \:|\: s_{1} \leq f\right\} = k \cdot \inf \left\{ \int_{a}^{b}{s_{1}} \:|\: s_{1} \leq f\right\}$$ but the lower integral of $g$ is not equal to the infimum of the set of the integrals of the step function below $g$.
Thanks for your insight