Let $\Theta\subset\mathbb{R}$ and consider the integral $$ \int_\Theta f(t) \left| \frac{1}{\sqrt{n}} g(x_n) t+ r_n(t) \right|dt, $$ where:
$\bullet$ $n$ is an integer sequence, the sequence of real numbers $x_n$ converges to $x$ as $n \to \infty$;
$\bullet$ $g$ is a real-valued function which is continuous in a neighborhood of $x$;
$\bullet$ $f$ is a positive, continuous function and $\int_\Theta |t|f(t)=c<\infty$;
$\bullet$ for any fixed $t$, $r_n(t)=O(1/n)$ and $\int_\Theta f(t)|r_n(t)|dt=O(1/n)$.
Under the above assumptions, would it be correct to write:
$$ \int_\Theta f(t) \left| \frac{1}{\sqrt{n}} g(x_n) t+ r_n(t) \right|dt= c \frac{|g(x_n)|}{\sqrt{n}}+O(1/n) $$ with equality sign? Of course, in light of triangle inequality, the above relation with "$=$" replaced by "$\leq$" holds true. Thus, I was wondering whether it is still valid with the equality sign.
Sure, it's valid. To show that the relation also holds with a $\geq$ sign, notice that $$\int_\Theta f(t) \left| \frac{1}{\sqrt{n}} g(x_n) t+ r_n(t) \right|\;dt \geq \int_\Theta f(t) \left| \frac{1}{\sqrt{n}} g(x_n) t\right| - f(t) |r_n(t)|\;dt \geq c \frac{|g(x_n)|}{\sqrt{n}}-O(1/n)$$ by the reverse triangle inequality.
Notice, however, that if $g(x_n) \to 0$ sufficiently rapidly (i.e. if $g(x) = 0$ and $x_n \to x$ fast enough) then the first term $c\frac{|g(x_n)|}{\sqrt{n}}$ may be asymptotically smaller than the $O(1/n)$ term. All of the proceeding analysis remains valid, but the asymptotic series you obtain is not as useful in applications as it would be if $g(\lim x_n) \neq 0$.