measure theory - Definition of a measurable function?

Are these two definitions of a real-valued, measurable function equivalent? ($(X, \Sigma, \mu)$ is the measure space.)

Definition 1: $f: X \to \mathbb{R}$ is said to be measurable if for all $\alpha \in \mathbb{R}$, $\{ x \mid f(x) > \alpha \} \in \Sigma$ (i.e., $f^{-1}(\alpha, \infty) \in \Sigma$)

Definition 2: $f: X \to \mathbb{R}$ is said to be measurable if, given the $\sigma$-algebra $\Sigma'$ on $\mathbb{R}$, $E \in \Sigma ' \implies f^{-1}(E) \in \Sigma$.

I am under the impression that Definition 1 is the definition of Lebesgue measurable, that is, $\Sigma '$ must be the $\sigma$-algebra of Lebesgue measurable sets, and $\mu = m$, Lebesgue measure, even though this is probably not true.

The reason I am asking this is that I am having trouble proving the statement that the pointwise limit of a sequence of Borel measurable functions is Borel measurable. I wasn't sure if I could just use the property we derived from Definition 1 that the pointwise limit of a sequence of measurable functions is measurable, or if Definition 1 only applies to Lebesgue measurable functions.

Answer

Definition 1 gives measurability of $f$ with respect to the Borel-$\sigma$-algebra on $\mathbb{R}$, i.e. the $\sigma$-algebra generated by the open sets. In contrast, Definition 2 defines the measurability of

$$f: (X,\Sigma) \to (\mathbb{R},\Sigma'),$$

i.e. we do not necessarily consider the Borel-$\sigma$-algebra on $\mathbb{R}$, but an arbitrary $\sigma$-algebra on $\mathbb{R}$. If $\Sigma' = \mathcal{B}(\mathbb{R})$, then both definitions are equivalent. This follows from the fact that the family $\{(a,\infty); a \in \mathbb{R}\}$ is a generator of $\mathcal{B}(\mathbb{R})$.