Why was this $\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx=\int_{[0,\infty)}(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$ written in the proof?
Does this is related to the measurability of the function $g_n(x)$?
I do not see how. Even so I think the answer is 'yes' because the remaining hypotheses of DCT are mentioned in the proof.
See the hypotheses of the dominated convergence theorem are
A sequence of measurable functions $f_n$
$g\in L(\mu)$
$|f_n|\le g$
$f_n\to f$ pointwise
Calculate $\lim_{n\to\infty}\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx$ .
$\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx=\int_{[0,\infty)}(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$ for each $n\in\mathbb{N}$. Let $g_n(x)=(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$ for each $n\in\mathbb{N}$ and $x\in[0,\infty) $.Then $g_n(x)\leq e^{-x}$ for each $x\in[0,\infty)$ and $n\in\mathbb{N}$. $\int_{[0,\infty)}e^{-x}dx$ exists. Moreover $g_n(x)$ converges to $g(x)=e^{-x}$ for each $x$. Then by Dominated Convergence theorem $\lim_{n\to\infty}\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx=\int_{[0,\infty)}e^{-x}dx$.