Fundamental Characterisation of Measureable Functions Explanation (Characterization) I am just starting learning about measure theory (what a great way to spend Christmas...!), and I am unclear on this following claim: Characterisation of Measurable Functions. It appears to be a very basic (and fundamental) result, but I'm not really sure how to show it. I'm only really interested in "$(2)$ and $(2')$". I'm sure that I can deduce the others from it.
This is what I've been thinking: Given measurable $f:E \rightarrow \Bbb R$, write $f = \lim f_n$, where $f_n$ is a simple function, which is known to be measurable. Then $f(x) \le \alpha$ just when when $x$ is in one of the sets in the simple function which has constant $\le \alpha$; eg
$$f_n = \sum_{k=1}^{K_n} c_k^{(n)} 1(A_k^{(n)})$$
where $1(\cdot)$ is the indicator function, satisfies $fn(x) \le \alpha$ just when $x \in A_k^{(n)}$ with $c_k^{(n)} \le \alpha$. [I'm not 100% sure how this translates into the limit...].
An explanation would be most appreciated! Thanks! :)
UPDATE
The definition for a measurable set that I have been given is just the elements of the $\sigma$-algebra. Now for a function:

Let $(E,\mathcal E)$ and $(G,\mathcal G)$ be measurable spaces. We say that a function $f:E \rightarrow G$ is measurable if $f^{-1}(A) \in \mathcal E$ for every $A \in \mathcal G$.

Using the definition given in a link on the page linked above, the main part of my question ($(1) \iff (2)$) is fairly trivial. I'm interested to know how the above (equivalent, I assume!) definition gives this result.
 A: (1) and (2) are the same since (2) is the definition of measurable in the reference above. So, we need only show $(2) \Longleftrightarrow (2')$.
Clearly (2) $\Rightarrow$ (2').
Let $L_t = \{x | f(x) \le t \}$.
Suppose (2') holds and $\beta \in \mathbb{R}$. Let $\alpha_n \in \mathbb{Q}$ be a decreasing sequence such that $\alpha_n \downarrow \beta$.
Since $L_{\alpha_n} \in \Sigma$ for all $n$, we have $\cap_n L_{\alpha_n} \in \Sigma$, and since $L_\beta = \cap_n L_{\alpha_n}$, we have $L_\beta \in \Sigma$. Hence (2) holds.
Some notes: When the range of a function is a topological space, there is an implied Borel $\sigma$-algebra generated by the open sets. The definition in the question defines measurability for a function, but often (especially when the range is $\mathbb{R}$) the range $\sigma$-algebra is implicitly taken to be the Borel $\sigma$-algebra.
Suppose ${\cal C}$ is a collection of subsets of $G$ and suppose $f^{-1}(C) \in {\cal E}$ for all $C \in {\cal C}$, then it is straightforward to show that $f^{-1}(C) \in {\cal E}$ for all $C \in \sigma({\cal C})$.
From this we see that if $f^{-1}((-\infty,\alpha]) \in {\cal E}$ for all $\alpha$, then since ${\cal B} = \sigma (\{(-\infty,\alpha]\}_\alpha)$ (with
${\cal B}$ being the Borel sets of $\mathbb{R}$), we see that $f$ is (Borel)
measurable. (The other direction is trivial.)
(Also see Definition of a measurable function?.)
