Let $a, b\in R$ be such that $a<b$ and $u\in L([a, +\infty);R)$. Prove that $$ v(t)=\frac{b-a}{(b-t)^2}u\left(\frac{b-a}{b-t}+a-1\right)\in L([a, b];R). $$ Thank you for all helping and guidance.
My effort. I intend to use the following theorem to solve the problem
Theorem 1. Suppose $g: [a, b)\rightarrow R$ is a function having finite derivatives almost everywhere on $[a, b]$ and $f : [a, \infty] \rightarrow R$ is a Lebesgue integrable function such that the range of g is contained in $[a, \infty]$. Then $f (g(x))$ g'(x) is Lebesgue integrable on [a, b].
Let $\displaystyle w(t)=\frac{b-a}{b-t}+a-1, t\in [a, b)$. Then $w: [a, b)\rightarrow R$ is a function having finite derivatives almost everywhere on $[a, b]$. Since $$ v(t)=-w^\prime(t)u(w(t)) $$ and $u\in L([a,+\infty);R)$ then by Theorem 1 we have $v\in L([a, b]; R)$.
Please help me to check this solution and where we can find a proof of the above Theorem 1.
Your idea (to make the change of variables) is correct. However, Theorem 1 is unlikely to be found in the literature (even if it's true), because there is little interest in having $f(g(x))g'(x)$ integrable without knowing that $$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt \tag{1}$$
And just having having finite derivatives almost everywhere" doesn't get you $(1)$. For example, let $g$ be the standard Cantor staircase and $f(x)\equiv 1$: then $(1)$ turns into $0=1$.
Your function $w$ is in fact a diffeomorphism. The change of variable formula for diffeomorphisms is, e.g., Theorem 2.47 in Folland's Real Analysis (and in most real analysis books).