Why is $\{x:f(x)\ge a\}=\bigcap_{n=1}^\infty\{x: f(x)\gt a-\frac{1}{n}\}$ and $\{x: f(x)\gt a\}=\bigcup_{n=1}^\infty\{x: f(x)\ge a+\frac{1}{n}\}$? It is known from measure theory that for all $a \in \mathbb{R}$, the following are equivalent:


*

*$f$ is measurable.

*The set $\{x : f(x) \ge a\}$ is measurable.

*The set $\{x : f(x) \gt a\}$ is measurable.
In virtually all textbooks, for the proof of the above is as follows:
$2 \rightarrow 3$: $\{x : f(x) \ge a\} = \bigcap_{n=1}^{\infty}\{x: f(x) \gt a - \frac{1}{n}\}$
$3 \rightarrow 2$: $\{x : f(x) \gt a\} = \bigcup_{n=1}^{\infty}\{x: f(x) \ge a + \frac{1}{n}\}$
Can someone clarify as to why those equalities above hold? I'm having difficult picturing it. For instance, why is:
$$\bigcup_{n=1}^{\infty}\{x: f(x) \ge a + \tfrac{1}{n}\} = \{x : f(x) \ge a + 1\} \cup\cdots \cup \{x: f(x) \ge a\}$$ equal to: $\{x: f(x) \gt a\}$ and NOT: $\{x: f(x) \ge a\}$?

Suppose that $x_0$ is a point at which $f(x)=a$. For which $n$ do you think  $x_0$ is going to be in the set
$$\{x:f(x)\geq a+\tfrac{1}{n}\} \quad ?$$
If it is not in any such set, how will it be in
$$\bigcup_{n=1}^\infty\{x:f(x)\geq a+\tfrac{1}{n}\}\quad ?$$

The fact that the question is phrased in terms of functions is not important; it is more generally true that, for all $a,b\in\mathbb{R}$,
$$a\geq b\iff a>b-\tfrac{1}{n}\text{ for all }n\in\mathbb{N}$$
and 
$$a>b\iff \text{ there is some }n\in\mathbb{N}\text{ such that }a\geq b+\tfrac{1}{n}.$$
