Are these subsets measurable?

Question

Let $f_n$ be a sequence of measurable functions on the measure space $(X, \Sigma)$ (so $\Sigma$ is a $\sigma$-algebra on $X$). How can I show that these two sets are measurable?

\begin{align} A & = \{x \in X : \textrm{the sequence $f_n(x)$ doesn't attain its upper bound}\} \\ B & = \{x \in X : \textrm{the sequence $f_n(x)$ is bounded}\} \end{align}

What have I tried? Well, \begin{align} B & = \bigcup_{k \ge 1} \{x : (\forall n) (|f_n(x)| < k)\} \\ & = \bigcup_{k \ge 1} \bigcap_{m \ge 1} \{x : |f_m(x)| < k\} \\ & = \bigcup_{k \ge 1} \bigcap_{m \ge 1} f_m^{-1}[(-k, k)]. \end{align}

So $B$ is a measurable set. How about $A$? Maybe it is not measurable?

Answer 1

Our functions are measurable, so are their differences. Let

$$A_{n,N}=\{x: f_n(x)\leq f_N(x)\}=\{x: f_n(x)-f_N(x)\leq 0\}.$$

The sets $\{A_{n,N}\}$ are measurable because they are level sets of measurable functions.

$f_n(x)$ attains its smallest upper bound if there exists an $N$ such that for all $n$ $f_n(x)\leq f_N(x)$, that is if

$$x\in \bigcup_{N=1}^{\infty}\bigcap_{n=1}^{\infty}A_{n,N}.$$

So, the set over which the sequence attains its smallest upper bound is measurable. The complement of the set above is $A$. So $A$ is measurable.