Let $G : T \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$, $T$ is an open set and suppose that $G$ is a $C^1$-diffeomorphism in the sense that $G$ is injective and $\det DG(x) \neq 0$ for all $x \in T$, where $DG(x)$ is the total derivative of $G$ at $x \in T$.
Let $B \subset T$ be a borel set of finite measure and suppose $U_n \subset T$ is a decreasing sequence of open sets, such that $B \subset U_n$ for all $n$, and $m(U_n \setminus B) < \frac{1}{2^n}$ for all $n$. In particular we have that $B \subset \cap_{n \geq 1} U_n = U$ and $m(U \setminus B) = 0$ where $m$ is the lebesgue measure.
Why is it necessarily the case that dominated convergence theorem can be applied to show that $$\lim_{n \rightarrow \infty} \int_{U_n} |\det DG(x)|dx = \int_{B} |\det DG(x)|dx$$?
This seems to imply to me that when $\int_{B} |\det DG(x)|dx < \infty$, $\int_{U_N} |\det DG(x)| dx <\infty$ for some $N \in \mathbb{N}$ (and thus for all $n \geq N$) but I have been unable to see why this is true.
Edit: For reference, this particular worry came from reading page 73 of Folland (Real Analysis : Modern Techniques and their Applications – 1984) and I have attached the section in that page which I did not understand the argument of : https://i.sstatic.net/a91Mt.jpg
Best Answer
It turns out that Folland has updated this argument, the corrected version can be found here https://sites.math.washington.edu/~folland/Homepage/oldreals.pdf, page 2 of the document (page 76).