Let $f_n : M \to \overline{\mathbb{R}}$ be a sequence of measurable functions where $M$ is a measure space.
Let
$$A := \{ x : (f_n(x)) ~\text{is Cauchy}\}.$$
Recall that a sequence $(f_n(x))$ is Cauchy if, and only if, for every $\epsilon > 0$, there exist $m_0,n_0$ natural numbers such that for all $(m,n)$ such that $n > n_0, m > m_0$ we have $$d(f_n(x),f_m(x)) < \epsilon.$$
I was asked to show that this set is measurable. So, I tried to write it as unions and intersections of measurable sets. My questions is:
Is $A$ equal to $$\bigcap_{k \in \mathbb{N}}\bigcup_{m_0,n_0 \in \mathbb{N}}\bigcap_{m,n}\left\{x : d(f_n(x),f_m(x)) < \frac{1}{k} ~\wedge~ m > m_0, n > n_0 \right\}?$$ If this is true, then
$$ \left\{x : d(f_n(x),f_m(x)) < \frac{1}{k} ~\wedge~ m > m_0, n > n_0 \right\} = d(f_n,f_m)^{-1}([-\infty,0))\cap 1^{-1}(m_0,\infty])\cap 1^{-1}([n_0,\infty),$$ so, $$\bigcap_{k \in \mathbb{N}}\bigcup_{m_0,n_0 \in \mathbb{N}}\bigcap_{m,n}d(f_n,f_m)^{-1}([-\infty,0))\cap 1^{-1}(m_0,\infty])\cap 1^{-1}([n_0,\infty),$$ and the claim follows.
Is this right?