When considering $\lim_{n \to \infty}\int_{0}^{g(n)}f_n(x)$, $g(n)\to \infty$ does $g$ matter when $f_n(x)$ is dominated?
I was solving the following problem $$\lim_{n\to \infty}\int_{0}^{e^n}\frac{x}{1+nx^2}$$ to my suprise it was not DCT or MCT, but simply integrating and calculating the limit. But then it hit me that the bound $e^n$ influences the result. If it was $n$ then the result would be 0, if it is $e^n$ it would be $1$. However, when the function inside is dominated it does not really matter how fast the bounds grow as pointwise $\chi_{[0,g(n)]}\to \chi_{[0,\infty)}$ What is the intuition behind this phenomena?
You answered your own question. If $|f_n| \leq h$ where $h$ integrable, $f_n \to f$ a.e. ptwise on $[0,\infty)$, and $g(n) \to \infty$, then $$\int_0^{g(n)} f_n = \int 1_{[0,g(n)]} f_n.$$ Then $1_{[0,g(n)]}f_n$ is dominated by $h$ still and $1_{[0,g(n)]} f_n \to f$ on $[0,\infty)$, so: $$\lim_n \int_0^{g(n)}f_n \to \int_0^\infty f.$$
The intuition is that you'll be eventually integrating over the whole line, so it doesn't matter how you get there.