Express $\{f\leq c\}$ as a countable intersection As stupid a question as it is, but I need to clarify this. 
When proving the equivalence of several definitions of a measurable function, it is usually shown that $\{f\leq c\} = \bigcap\limits_{n=1}^{\infty}{\{f<c+\frac{1}{n}}\}$ is a Lebesgue measurable set etc. 
Intuitively, I have always understood why this set can be written as a countable intersection, but how is this property derived rigorously? 
Are such properties considered in real analysis or set theory?
 A: If $f(x) \leq c$ then $f(x) <c+\frac 1 n$ for all $n$ so LHS is contained in RHS. If $f(x) <c+\frac 1 n$ for all $n$, then, taking limits as $n \to \infty$ we gat $f(x) \leq c$. Hence RHS is contained in LHS. This as rigorous as it can be.
A: In any Archimedean field $\mathbb{K}$, you have no infinitesimal (or infinite, for that matter) elements.  This means that for every positive $x \in \mathbb{K}$ there is atleast one natural number $n$ such that $nx \geq 1$ (understood as summing $x$ $n$-times with itself), or - being a field - $x \geq \frac{1}{n}$ (where $n$ is understood to be the sum of $1$ $n$-times with itself and $\frac{1}{n}$ its inverse). This completely characterizes any positive number - so you can dually define a negative (or null) element $y$ to be one such that $y < \frac{1}{n}$ holds for any $n$, which means that if you have a set of the form 
\begin{equation}
\{x \in \mathbb{K}: x < 0\}
\end{equation} you can rewrite is as the countable union of sets of the form {$x \in \mathbb{K}$: $x < \frac{1}{n}$}.
