Definition of a measurable function?
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 $\mathcal{B}(\mathbb{R})$.
Definition 2 doesn't really make sense. Or, more properly it is not equivalent to def 1 nor useful.
In measurability, you want a sigma algebra (i.e., a measurable space) in your domain, but a topological one in your codomain. Otherwise, if you don't have a notion of convergence in your codomain, you cannot really define a useful integral.
Definition 2, to be equivalent with definition 1, should be that given $E$ open, $f^{-1}(E)\in \Sigma$.
Definition 1 follows from definition 2 (assuming $\Sigma'$ is the Borel sigma algebra) because of the following lemma.
Suppose $(X,\Sigma)$ and $(X',\Sigma')$ are two measurable spaces, and suppose that the $\sigma$-algebra $\Sigma'$ is generated by the family of sets $\Pi$. Then $f : X \rightarrow X'$ is $\Sigma/\Sigma'$ measurable if (and only if, trivially) $f^{-1}(E) \in \Sigma$ for all $E \in \Pi$.
(Proof hint: $\{ E : f^{-1}(E) \in \Sigma\}$ is a $\sigma$-algebra containing $\Pi$.) Apply the lemma to the family $\{(-\infty,a) : a \in \mathbb{R}\}$, which generates the Borel $\sigma$-algebra. The Lebesgue $\sigma$-algebra is slightly larger; it is the completion of the Borel $\sigma$ algebra.
As for your question, here are two hints:
$\{\sup_j f_j \leq a\} = \cap_{j=1}^\infty \{f_j \leq x\}$.
$\limsup_{j\rightarrow\infty} f_j = \inf_n \sup_{j \geq n} X_j.$
Prove the same relation for $\liminf$ and conclude that both $\limsup f_j$ and $\liminf f_j$ are Borel-measurable. If $\lim_{j\rightarrow\infty} f_j$ exists, then ...