Sequence of measurable functions (lim sup)

Let $$X$$ be a set, $$\mathcal{A}$$ a sigma-algebra on $$X$$ and $$\mathcal{B}$$ the Borel algebra on $$\mathbb{R}$$.

Let $$\lbrace f_n:(X,\mathcal{A})\to (\mathbb{R},\mathcal{B})\rbrace_{n\in\mathbb{Z_+}}$$ be a sequence of measurable functions, such that a $$K>0$$ exists with $$|f_n(x)| for all $$n \in \mathbb{Z_+}, x\in X$$.

How to show, that $$f:(X, \mathcal{A})\to(\mathbb{R},\mathcal{B}), x \mapsto \limsup\limits_{n\to\infty}f_n(x)$$ is also measurable?

I used the following steps:

$$\lbrace \sup\limits_{n\in \mathbb{N}}f_n < K\rbrace=\bigcap\limits_{n=1}^{\infty} \lbrace f_n so the supremum is measurable.

Also, $$\limsup\limits_{n\to\infty}f_n$$=$$\inf\limits_{n\in\mathbb{N}}\sup\limits_{l\geq n}f_l$$ so the limit superior is measurable.

Can it be done like that or is there another way to show it right?

$$\{x:\ sup_{n\in \Bbb N}f_n\ (x)> \alpha\}=\bigcup_{n=1}^{\infty}\{x:\ f_n(x)>\alpha\}$$ for each $$\alpha\in \Bbb R$$. Hence $$sup\ f_n$$ is a measurable function.
Now since $$inf_{n\in \Bbb N}\ g_n=-sup_{n\in \Bbb N}\ (-g_n)$$ we can say $$inf_{n\in \Bbb N}\ g_n$$ is also measurable whenever $$g_n$$ are also measurable.
Finally $$lim\ sup_{n\rightarrow \infty}\ f_n=inf_{n\in \Bbb N}(sup_{i≥n}\ f_i)$$. So that $$lim\ sup_{n\rightarrow \infty}\ f_n$$ is also measurable.