Let $(\Theta,d)$ be a metric space, $\mathcal{X}\subset\mathbb{R}^n$ and $f:\Theta\times \mathcal{X}\to\mathbb{R}$. Define $X$ as a random variable in $\mathcal{X}$. Suppose that the mapping $\theta\mapsto E[f(\theta,X)]$ is continuous.
For a fixed $\theta\in\Theta$ define the subsets $B_1(\theta),B_2(\theta),...$ as $B_j(\theta)=\{\theta':d(\theta,\theta')< 1/j\}$. Therefore, $B_1(\theta)\supset B_2(\theta)\supset\cdots $ and $B_j(\theta)\downarrow\{\theta\}$ as $j\to\infty$.
Is it true that $$ \lim_{j\to\infty}E\left[\sup_{\theta'\in B_j(\theta)}f(\theta',X) - \inf_{\theta'\in B_j(\theta)}f(\theta',X)\right] = E[f(\theta,X)]-E[f(\theta,X)]=0\quad ? $$ My approach was by defining $\theta_j^U$ and $\theta_j^L$ such that $$ f(\theta_j^U,X) = \sup_{\theta'\in B_j(\theta)}f(\theta',X),\quad f(\theta_j^L,X) = \inf_{\theta'\in B_j(\theta)}f(\theta',X). $$ So by continuity, we have $$ \lim_{j\to\infty}E\left[\sup_{\theta'\in B_j(\theta)}f(\theta',X)\right]=\lim_{j\to\infty}E[f(\theta_j^U,X)]=E[f(\lim_{j\to\infty}\theta_j^U,X)]=E[f(\theta,X)]. $$ The weak arguments are that $\theta_j^U$ might not be well defined and that $\theta_j^U$ not necessarily converges to $\theta$ a.s.
Can anybody provide some insights? Is the affirmation true? If not, under what conditions is it true?
I don't think this is true. If we take $\Theta = [-1,1]$, $X = \pm 1$ with probability $\frac 12$, and define $$f(\theta, x) = \begin{cases} x & \theta \le 0 \\ -x & \theta > 0\end{cases}$$ then $\mathbb{E}[f(\theta,X)] = 0$ so in particular $\theta \mapsto \mathbb{E}[f(\theta,X)]$ is continuous. However, if we fix $\theta = 0$ then $\sup_{\theta' \in B_j(0)} f(\theta',X) = 1$ and $\inf_{\theta' \in B_j(0)} f(\theta',X) = -1$ for every $j > 0$ so $$\lim_{j \rightarrow \infty} \mathbb{E}\left[\sup_{\theta' \in B_j(0)} f(\theta',X) - \inf_{\theta' \in B_j(0)} f(\theta',X)\right] = 2.$$
I'm not sure exactly what the conditions you need are, but I would guess that at the very least you need $f(\theta,x)$ to be continuous in $\theta$.
Edit: If you have $\Theta$ is locally compact, $\theta \mapsto f(\theta,x)$ is continuous, and $\mathbb{E}[\sup_{\theta \in \Theta} |f(\theta,X)|] < \infty$, then the dominated convergence theorem gives
$$\lim_{j \rightarrow \infty} \mathbb{E}\left[\sup_{\theta' \in B_j(\theta)} f(\theta',X) - \inf_{\theta' \in B_j(\theta)} f(\theta',X)\right] = \mathbb{E} \left[\lim_{j \rightarrow \infty}\left( \sup_{\theta' \in B_j(\theta)} f(\theta',X) \right)- \lim_{j \rightarrow \infty}\left( \inf_{\theta' \in B_j(\theta)} f(\theta',X)\right)\right] = \mathbb{E}[f(\theta,X) - f(\theta,X)] = 0$$