Let $(f_n(x)):\Omega\rightarrow \mathbb{R}$ be a sequence of measurable functions, the sets
$$A=\{x \in \Omega: (f_n(x)) \text{ converges to a real number}\}$$
$$B=\{x \in \Omega: (f_n(x)) \text{ converges to a rational number}\}$$
$$C=\{x \in \Omega: (f_n(x)) \text{ converges to an irrational number}\}$$
are measurable?
This type of question is answered in many places, but there are a few confusions that I have after reading.
For example: The set that $(f_n)$ converges to a real number is measurable, it says
$\{x \in \Omega: (f_n(x)) \text{ converges to a real number}\}$ is equivalent to $\{x \in \Omega: (f_n(x)) \text{ is Cauchy}\}$.
Here are my confusions:
- I understand that $\mathbb{R}$ is complete, but why do we work with Cauchy instead of convergence itself? Here: $$A=\{x\in\Omega: \forall k \in \mathbb{N},\, \exists N \in \mathbb{N} \text{ s.t. } \forall n>N,\, |f_n(x)-r_x| < 1/k\}$$ where $r_x\in\mathbb{R}$ is the pointwise limit of $(f_n(x))$, $x\in \Omega$. Then $$A = \bigcap_{k\in\mathbb{N}}\bigcup_{N\in\mathbb{N}}\bigcap_{n > N} \{x\in\Omega:|f_n(x)-r_x|<1/k\}$$Since $r_x$ are constants, therefore the set $\{x\in\Omega:|f_n(x)-r_x|<1/k\}$ is measurable, and therefore $A$ is measurable. (right???) I don't see why we can't work with convergence directly.
- Suppose what they did in the link is correct, why can we ignore what number that $(f_n(x))$ converges to? What if $f_n(x)$ converges to a rational (set $B$), or an irrational (set $C$)? The arguments are the same? And does that mean $B=C$ if we only need $(f_n(x))$ to be Cauchy? It does not feel right.
Thanks for any insights and help.
Best Answer
For a different perspective, let $g = \liminf_{n \to \infty}f_n$, $h = \limsup_{n \to \infty}f_n$. It is well known that $g$ and $h$ are both measurable. We have $$A = (h - g)^{-1}(\{0\}),$$ $$B = A \cap g^{-1}(\mathbb{Q}),$$ $$C = A \cap g^{-1}(\mathbb{R} \setminus \mathbb{Q}),$$ all measurable.