I had this excercise in measure theory where I had to show that certain sets are measurable and I realized there was some mechanical procedure going on. Here is the question:
Let $f_n:X\to \mathbb{R}$ be measurable functions. Show the measurability of the following sets:
$A = \{x:f_n(x)\to \infty\}$
$B = \{x:f_n(x)\to -\infty\}$
$C = \{x:f_n(x)\to \exists \lim_{n\to \infty}f_n \text{ finite > limit}\}$
Here's my solution:
$A= \bigcap_{M>1} \bigcup_{N>1} \bigcap_{n>N}\{x:f_n(x)>M\} \implies$Measurable
$B= \bigcap_{M>1} \bigcup_{N>1} \bigcap_{n>N}\{x:f_n(x)<-M\} \implies$Measurable
$C= \bigcap_{M>1} \bigcup_{N>1} \bigcap_{n>m>N}\{x:|f_m(x)-f_n(x)|<\frac{1}{M}\} \implies$Measurable
Is it generally true that you can translate anything defined by quantifiers to unions and intersection using the rules $\exists \to \bigcup$ and $\forall \to \bigcap$?
Unfortunately, original poster isn’t around to explain how broad extent of the problem should be considered. Literally “anything defined by quantifiers” in mathematics can be translated to set operations only where the theory uses semantics of set theory. Namely, if we have logical formula
(quantifiers on $k_1,\ldots, k_n$ in said order) $P(x, k_1,\ldots, k_n)$
where $x$ is assumed to belong to certain domain of discourse $D$ (this clause mitigates set-theoretical troubles with “proper classes”) and $P$ is a proposition (predicate) depending on said variables through set membership relation, i.e. expressions “variable ∈ some_set ”, then we can, in principle, equivalently transform this formula to “$x∈S\,$” where $S⊂D$ is certain set constructed from original formula. Of course, each “some_set ” can depend on (other) variables of the $k$ series.
How “mechanically” can it be done? If $P$ contains only “$x∈X(\cdots)$ where $X(\cdots)⊂D$ ” things, but not set membership conditions on $k_1,\ldots, k_n\,$ (other than their domains of discourse, see below), then fully. Rules of transformation follow: $$ x∈X_1\ ∧\ x∈X_2 \quad\text{becomes}\quad x\ ∈\ X_1∩X_2 \\ x∈X_1\ ∨\ x∈X_2 \quad\text{becomes}\quad x\ ∈\ X_1∪X_2 \\ ¬(x∈X_1) \quad\text{becomes}\quad x\ ∈\ D\setminus X_1 \\ ∀k_i∈K_i:\ x∈X(\ldots,k_i) \quad\text{becomes}\quad x∈\bigcap_{k_i∈K_i}X(\ldots,k_i) \\\text{(assuming }K_i\text{ isn’t empty)}\\ \ \\ ∃k_i∈K_i:\ x∈X(\ldots,k_i) \quad\text{becomes}\quad x∈\bigcup_{k_i∈K_i}X(\ldots,k_i) $$
Applying these rules to inner parts of the formula and gradually moving $k_n,\ldots, k_1$ from quantifiers to operations on set families, we can eventually translate all connectives and all quantifiers. We see, these rules can deal with situation where domain $K_i$ of $k_i$ depends on previous variables ($k_1,\ldots, k_{i-1}$).
What can we do if our formula contains also some $R(k_1,\ldots,k_i)$ propositions? Each time we face
$Qk_i∈K_i:$ (some formula on membership of $x$ in many sets and predicates on $k_1,\ldots,k_i$),
where Q is a quantifier (either ∀ or ∃), then we should convert the formula to either conjunctive(∀) or disjunctive(∃) normal form. We know that $$ ∀k: P_1(k)∧P_2(k)\quad\Leftrightarrow\quad(∀k: P_1(k))∧(∀k: P_2(k)) \\ ∃k: P_1(k)∨P_2(k)\quad\Leftrightarrow\quad(∃k: P_1(k))∨(∃k: P_2(k)) $$ and all propositions independent of $k_i$ can be extracted out of the quantifier, and eventually we should learn how to deal only with: $$ ∀k_i∈K_i:\ x∈X(\ldots,k_i)∨R(\ldots,k_i)\quad(1)\\ ∃k_i∈K_i:\ x∈X(\ldots,k_i)∧R(\ldots,k_i)\quad(2)\\ $$ Obviously, equivalent forms are $$ x∈D\ \text{for }(1)\ \text{if}\ ∀k_i∈K_i:\ R(\ldots,k_i)\\ x∈\bigcap_{k_i∈\{q∈K_i\ |\ ¬R(\ldots,q)\}}X(\ldots,k_i)\ \text{for }(1)\text{ otherwise}\\ x∈\bigcup_{k_i∈\{q∈K_i\ |\ R(\ldots,q)\}}X(\ldots,k_i)\ \text{for }(2) $$ The only thing we can’t do “mechanically” in this way is checking $∀k_i∈K_i:\ R(\ldots,k_i)$ each time we eliminate disjunction under a universal quantifier.
I hope it addresses original poster’s problem sufficiently.