Assume we have some fixed (finite, may be a density, but not necessarily) function $f(x)$
and some other function $g(\theta,x)$ such that for fixed $x$, $g(\theta,x)$ is in $O(\theta^m)$.
Furthermore, assume a random variable $\delta_n$ such that $\delta_n \overset{\mathbb{P}}{\rightarrow} 0 $. Is this enough to state that $\int f(x) g(\delta_n,x) dx \overset{\mathbb{P}}{\rightarrow} 0$ ?
The specific situation where I have seen this argument was that $f(x) = p(\theta,x)$ for some fixed $\theta$ and $g(\theta,x)$ is the rest term of the taylor expansion of the respective log-likelihood $\ln p(\theta,x)$ around $\theta$, and it was not explained why this works. (And i couldn't figure out how to see it)
Partial answer to my question:
Assume we can write $$g(\theta,x) = \theta^m h(x)$$ for some function $h$, which will be true in the above case of a Taylor expansion at $0$.
Then we have that $$\int f(x) g(\theta,x) dx = \theta^m \int f(x)h(x)dx$$
As long as $|\int f(x) h(x) dx| < \infty $, this is continuos in $\theta$ at $0$ , thus the claim follows by the continuos mapping theorem.