Just a quick question on the structure of the following proof. My book defines a function $f$ to be measurable on $E \subset \mathbb{R}^d$, if for all $a \in \mathbb{R}$ the set
$$f^{-1}([-\infty , a)) = \{x \in E: f(x) < a\}$$
is measurable.
Next it is claimed that we have an equivalent definition if we use the inverse image of closed intervals (they use an iff statement). The proof is obtained by seeing that in one direction we have
$$\{f \leq a\} = \bigcap_{k = 1}^\infty \{f < a + 1/k\} \tag{1}$$
and in the other direction
$$\{f < a\} = \bigcup_{k = 1}^\infty \{f \leq a - 1/k\} \tag{2}$$
Noting that for $(1)$ a countable intersection of measurable sets is measurable.
Okay, so I take the proof to mean that for $(1)$ we are proving the following: If $\{f < a\}$ is measurable, then $\{f \leq a\}$ is measurable. Then we show that because the right hand side is a measurable set for all $k$ by definition, and its countably infinite intersection is the left side.
For $(2)$ I believe that we are showing the same thing, essentially in reverse.
Really what I'm asking is: is the bolded reason above correct?