The proof of general strong law of large numbers

Question

Let the function $f\colon \mathbb R^n \times \Xi \to \mathbb R$ be measurable, $f(\cdot, \xi)$ for $\xi$ fixed be lower semicontinuous. Supose that there exists an $\mathbf P$-integrable ($\mathbf P$ is a probabilistic measure) function $a(\xi)$ such that $$f(x, \xi) \ge a(\xi)$$ holds true for all $x\in\mathbb R^n$. Let $$F_n(x,\xi_1,\ldots,\xi_n) = \frac 1n \sum_{k=1}^n f(x, \xi_k).$$ Let $\{x_i\}$ be a sequence such that $$\lim_{i\to\infty} x_i = x_0.$$ It is known (the standard stong law of large numbers) that for all $i\in\mathbb N$ $$\lim_{n\to\infty}F_n(x_i,\xi_1,\ldots,\xi_n) = (Ef)(x_i)$$ for all sequences $\{\xi_n\}\in \Theta$, where $\Theta$ is a set of full measure, $E$ is the expectation. Let a sequence $\{\xi_n\}\in \Theta$ be fixed. Also, we suppose that $$(Ef)(x_i)\to (Ef)(x_0).$$ Is this true that there exists a subsequnce, say $\{y_n\}$, of $\{x_n\}$ such that $$\lim_{n\to\infty}F_n(y_n,\xi_1,\ldots,\xi_n) = (Ef)(x_0)?$$ The authors of the paper [Artstein Z., Wets R.J.-B. Consistency of minimizers and the SLLN for stochastic programs // Journal of Convex Analysis. 1996. V. 2. P. 1-17.] write that this is clear, but I don't understand this. One of the convergences should be uniform, and this is not obvious. This property is used to prove the general strong law of large numbers for stochastic programs.