Uncountable union of sets I proved that that $\left\{\lim_n x_n = \ell \right\}\in \mathcal{F}$ where $\mathcal{F}$ is a sigma algebra , $\left\{X_n\right\}_{n\in \mathbb{N}}$ is a random variable.
We also have to show that 
$\bigcup_{\ell\in \mathbb{R}}\left\{\lim_n x_n = \ell\right\} \in \mathcal{F}$ which seems to be true if we think that 
$$\left\{\lim_n x_n = \ell_1\right\}\cup \cdots \cup \left\{\lim_n x_n = \ell_i \right\}\cup\cdots$$
where all $\left\{\lim_n x_n=\ell_i\right\}\in\mathcal{F}$
So from definition of sigma algebra the union of all that kind of sets should be in $\mathcal{F}$,but what makes it uncertain is that the union $\bigcup_{\ell\in\mathbb{R}}$ is uncountable.
Any idea how to surpass that or my intuition is right ??
 A: Use that $\exists \ell : X_n(\omega) \to \ell$ iff $X_n(\omega)$ is a Cauchy sequence in $\mathbb{R}$, by completeness. The latter we can state in quantifiers over $\mathbb{N}$ as $$\forall n: \exists m: \forall k_1,k_2 \ge m: |X_{k_1}(\omega) - X_{k_2}(\omega)| < \frac{1}{n}\text{.}$$ Now write this in a way which makes it clear that this set of $\omega$ is in $\mathcal{F}$.
A: To see that the approach with an uncountable union typically will not succeed, consider the following example: Suppose that $\Omega=\mathbb{R},$ $\mathcal{F}=\mathcal{B}(\mathbb{R}),$ and let $X_{n}(\omega)=\omega$ for all $n\in\mathbb{N}.$ Then $\{\lim_{n}X_{n}=\ell\}=\{\ell\}=\bigcap_{n\geq1}(\ell-1/n,\ell+1/n)\in\mathcal{F},$ since $\mathcal{B}(\mathbb{R})$ is the $\sigma$-algebra generated by the open intervals.
Now let $\mathcal{N}$ be your favorite nonmeasurable subset of $\mathbb{R}$. Then $$\bigcup_{\ell\in\mathcal{N}}\{\lim_{n}X_{n}=\ell\}=\bigcup_{\ell\in\mathcal{N}}\{\ell\}=\mathcal{N},$$ which by assumption is nonmeasurable.
