Below is a statement of a question I've been struggle to understand; I also attached a proof below the statement. I feel that it's a very incomplete and inaccurate proof right off the bat. If anyone can possible hint at how to adjust my first case, the rest will follow from that. However, as written, I'm not comfortable saying I've satisfied the result.
The question is:
Let $f \colon \mathbb{R} \to \mathbb{R}$ be integrable and $a \neq 0$. Then $g(x)=f(ax)$ is integrable and $\int f = |a| \int g$.
We break this proof into cases, each of which depends on how $f$ is defined.
(i) Suppose that $f(x) = \chi_A(x)$, where $A$ is some measurable set of finite measure. Then $$ g(x) = \begin{cases} 1, &\text{if } ax \in A\\ 0, &\text{if } ax \not \in A. \end{cases} $$ Then \begin{align*} \int |g(x)| &= \int |f(ax)|\\ &= \int \begin{cases} 1, &\text{if } ax \in A\\ 0, &\text{if } ax \not \in A. \end{cases}\\ &= \underbrace{ \int \frac{1}{|a|} \begin{cases} 1, &\text{if } x \in A\\ 0, &\text{if } x \not \in A. \end{cases}}_{\text{ This line is my main concern}}\\ &= \frac{1}{|a|} \int f, \end{align*} verifying the result for the first case.
(ii) Suppose that $f$ is a nonnegative, integrable simple function with canonical representation given by $f(x) = \sum_{i=0}^{n} e_i \chi_{E_i}$. Then \begin{align*} \int |g(x)| &+ \int |f(ax)|\\ &= \int \sum_{i=0}^n e_i \chi_{E_i}\\ &= e_o \int \begin{cases} 1, &\text{if } ax \in E_0\\ 0, &\text{if } ax \not \in E_0. \end{cases} + \cdots + e_n \int \begin{cases} 1, &\text{if } ax \in E_n\\ 0, &\text{if } ax \not \in E_n. \end{cases}\\ &= \frac{e_o}{|a|} \int \begin{cases} 1, &\text{if } x \in E_0\\ 0, &\text{if } x \not \in E_0. \end{cases} + \cdots + \frac{e_n}{|a|} \int \begin{cases} 1, &\text{if } x \in E_n\\ 0, &\text{if } x \not \in E_n. \end{cases}\\ &= \frac{1}{|a|} \sum_{i=0}^{n} e_i m(E_i) = \frac{1}{|a|} \int f, \end{align*} verifying the result for this case.
(iii) For this case, suppose that $f$ is an arbitrary nonnegative integrable function. Then, there exist a sequence of nonnegative, integrable simple functions $(\varphi_n)$, whose terms have canonical representation $\varphi(x) = \sum_{i=0}^m e_i \chi_{E_i}$, increasing pointwise everywhere to $f$. Then $$ \int |g(x)| = \int |f(ax)| = \int f(ax) = \int \lim_{n \to \infty} \varphi_n(ax). $$ With how we've defined the sequence, we may apply the Monotone Convergence Theorem the the fine equality to get $$ \int \lim_{n \to \infty} \varphi_n(ax) = \lim_{n \to \infty} \int \varphi_n(ax) = \frac{1}{|a|} \lim_{n \to \infty} \int \varphi_{n}(x), $$ where the last equality holds from the preceding case. With a second application of the Monotone Convergence Theorem, we see that $$ \int |g(x)| =\frac{1}{|a|} \lim_{n \to \infty} \int \varphi_{n}(x) = \frac{1}{|a|} \int \lim_{n \to \infty} \varphi_{n}(x) = \frac{1}{|a|} \int f, $$ completing this case.
(iv) Finally, let $f$ be any integrable function. Then we can reduce to case two by writing $f = f^+-f^-$, the difference of its positive and negative parts, which is the different of nonnegative integrable function, completing the proof.