Correcting a proof regarding Lebesgue integral.

Question

The question and a proof I found on the internet is given below:

Suppose that $g$ is integrable over $\mathbb{R},$ let $a, b \in \mathbb{R}$ and suppose that $a \neq 0$. Prove or disprove: the function $h(x) = g(a x + b)$ is integrable over $\mathbb{R}.$

the statement is correct and here is the proof. Suppose that $g$ is integrable over $\mathbb{R},$ Then $\int_{\mathbb{R}} g(u) \, du < \infty $ by definition. Now, define $h(x) = g(a x + b)$ for $a \neq 0.$\

Let $u = a x + b ,$ then $du = a \,dx,$ so $dx = \frac{du}{a}.$ Also, $\int_{\mathbb{R}} h(x) \,dx = \int_{\mathbb{R}} g(a x + b) \, dx = (1/a ) \int_{\mathbb{R}} g(u) \, du,$ which is $< \infty$ by the given, so $\int_{\mathbb{R}} h(x) \, dx $ is finite, so $h$ is integrable.

I doubt that this proof is correct from one of my previous questions , could anyone point out a hint for the correct proof?

Answer 1

HINT

Begin with proving this for every function $1_A(x)$ where $A$ is measurable with finite measure.

Then prove this for non-negative functions with approximation by an increasing sequence of non-negative simple functions and by using the monotone convergence theorem.

Then for the general case,apply the previous case to $f^-,f^+$

EDIT

Since you have proved the first two steps for simple functions then for non-negative $f$:

Exists an increasing sequence $h_n$ of non-negative simple functions such that $h_n \to f$ pointwise thus $g_n(ax+b) \to f(ax+b)$

We have by monotone convergence:

$$\int_{\Bbb{R}}f(x)dx=\lim_n\int_{\Bbb{R}}h_n(x)dx=\lim_n\int_{\Bbb{R}}h_n(ax+b)dx=\int_{\Bbb{R}}f(ax+b)dx$$

Continue from here..