Sequence of Measurable Functions (Unsigned and Complex-Valued) I am having several difficulties in solving the following problem about measurable functions:
Let $(X,\mathcal{B})$ be a measurable space.
If $f_n : X \to [0,+\infty]$ are a sequence of measurable functions that converge pointwise to a limit $f:X \to [0,+\infty]$, then show that $f$ is also measurable. Obtain the same claim if $[0,+\infty]$ is replaced by $\mathbb{C}.$ 
The given definition of a measurable function is analogous to that of a continuous function in Topology, according to Terrence Tao which is the author of the textbook I am using:
Let $(X,\mathcal{B})$ be a measurable space, and let $f: X \to [0,+\infty]$ or $f:X \to \mathbb{C}$ be an unsigned or complex-valued function. We say $f$ is measurable if $f^{-1}(U)$ is $\mathcal{B}-$measurable for every open subset $U$ of $[0,+\infty]$ or $\mathbb{C}$.
This is my failed attempt to a solution since I think that I am just writing definitions and not actually proving anything.
$(X,\mathcal{B})$ is a measurable space if $\mathcal{B}$ is a $\sigma-$algebra of subsets of $X$. We wish to show that $f:X\to [0,+\infty]$ is measurable if $f_n : X \to [0,\infty]$ are a sequence of measurable functions converging pointwise to $f$. Since we are in a $\sigma-$algebra, then we get $\forall E_1,E_2 \cdots \in \mathcal{B}, \quad \cup_{n=1}^{\infty}E_n$, call this countable union $U$. 
Because the $f_n$ sequence are measurable functions, then $f_n^{-1}(U)$ is $\mathcal{B}-$measurable for every open subset $U$ of $[0,+\infty]$. Our goal is to use this to show that $f$ is measurable. 
Here is where I am stuck. How could I go to the next step from here? Additionally, how would this solution extend to $\mathbb{C}$?
Update: I have been reviewing some lecture notes and the textbook and noticed that some of the proofs make a jump from using the quantifiers $\exists$ and $\forall$ to switch with $\cup$ and $\cap$ respectively. 
Just in case anyone is interested in reading up this problem, it is coming from Terrence Tao's Introduction to Measure Theory. This is Exercise 1.4.29 (vi). 
Here is the link to free pdf version: 
http://terrytao.files.wordpress.com/2011/01/measure-book1.pdf
 A: An easy way the result, is to see that convergence implies that
$$
f(x) = \lim_{n \to \infty}f_n(x) = \limsup_{n \to \infty } f_n (x).
$$
That is, $f(x) = \inf_m \sup_{k>m} f_k (x)$, from which the result follows since the supremum and infimum of countably many measurable functions is measurable. 
A: Since $f_n$ converges pointwise to $f$, we have $f(x)=\lim_{n \to \infty}f_n(x) = \lim \sup_{n\to \infty}f_n(x) = \inf_{N>0}\sup_{n\geq N}f_n(x)$.
This implies that for some $\lambda$, the set $\{x \in [0,+\infty]:f(x)> \lambda\}$ is equal to $\cup_{M>0}\cap_{N>0}\{x\in [0,+\infty]: \sup_{n \geq N} f_n(x) > \lambda + \frac{1}{M}\}$ outside of a set with measure zero. Actually, this set is 
$\cup_{M>0}\cap_{N>0}\cup_{n \geq N}\{x \in [0,+\infty]: f_n(x)> \lambda + \frac{1}{M}\}$ outside of a set with measure zero. Because each $f_n$ is measurable, the set $\{x\in [0,+\infty]: f_n(x)> \lambda + \frac{1}{M}\}$ is measurable. Since countable unions and intersections are measurable in $\sigma-$algebras, which is what is given since $(X,\mathcal{B})$ is a measurable space, then $f$ is measurable. 
This is what I have so far, although I am unsure if I can simply change the domain from $[0,+\infty]$ to $\mathbb{C}$ and then use the same argument. Any suggestions would be greatly appreciated since I have spent a lot of time on this problem already. 
