Given a measure space (X,X), we define a function f on X to $\mathbb{R}$ to be X-measurable if for every real number $\alpha$ the set $\{x \in X : f(x) > \alpha\}$ is in X.
Now one can show that the statements:
For every $\alpha \in \mathbb{R}$: $A_{\alpha} = \{ x \in X : f(x) > \alpha\} \subset \textbf{X}$
For every $\alpha \in \mathbb{R}$: $C_{\alpha} = \{ x \in X : f(x) \geq \alpha\} \subset \textbf{X}$
are equivalent.
To prove the first inclusion one should be able to see that $C_{\alpha} = \cap_{n=1}^{\infty} A_{\alpha - \frac{1}{n}}$
But I fail to see that, since even in the limit, $A_{\alpha - \frac{1}{n}}$ still contains elements x for which $f(x) > \alpha$ by the definition of $A_{\alpha}$. How does this take make sure that elements are included for which equality holds?
Because if $x\in C_\alpha$, that is, if $f(x)\geqslant\alpha$, then$$(\forall n\in\mathbb{N}):f(x)>\alpha-\frac1n,\tag1$$and therefore$$x\in\bigcap_{n\in\mathbb{N}}A_{\alpha-\frac1n}.\tag2$$On the other and, if $(2)$ holds, then we have $(1)$ and therefore $f(x)\geqslant\alpha$. In other words, $x\in C_\alpha$.