# 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).

• Perhaps start by showing that $f^{-1}((c,\infty))$ is measurable, by comparing it to the $f_n((c,\infty))$? – Greg Martin Oct 5 '14 at 1:22
• If I compare the pre-image to the pre-image of the sequence of functions, then could I just say that since it converges pointwise to $f$, then it must equal to $f$? I feel like this would not be sufficient. – Jamil_V Oct 5 '14 at 5:28

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.
• Ok, and would this also apply for the case when we move to $\mathbb{C}$? If possible, could you include some more details in the solution? – Jamil_V Oct 5 '14 at 5:27
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.