21.3 Real-Valued Measurable Functions
Lemma 21.3.1.label Let $(X, \cm)$ be a measurable space and $f: X \to \real$, then the following are equivalent:
- (1)
$f$ is $(\cm, \cb_{\ol \real})$-measurable.
- (2)
$f$ is $(\cm, \cb_{\real})$-measurable.
Proof. (1) $\Rightarrow$ (2): $\cb_{\real}\subset \cb_{\ol \real}$.
(2) $\Rightarrow$ (1): For any $E \subset \ol{\real}$, $f^{-1}(E) = f^{-1}(E \cap \real) \in \cm$.$\square$
Proposition 21.3.2.label Let $(X, \cm)$ be a measurable space, $\seq{f_n}$ be $(\cm, \cb_{\ol \real})$-measurable functions, then the following functions are $(\cm, \cb_{\ol \real})$-measurable:
- (1)
$F = \sup_{n \in \natp}f_{n}$.
- (2)
$f = \inf_{n \in \natp}f_{n}$.
- (3)
$G = \limsup_{n \to \infty}f_{n}$.
- (4)
$g = \limsup_{n \to \infty}f_{n}$.
- (5)
$\limv{n}f_{n}$ (if it exists).
In addition, if the above functions are $\real$-valued, then they are $(\cm, \cb_{\real})$-measurable.
Proof. (1): Let $\alpha \in \real$, then for any $x \in X$, $F(x) > \alpha$ if and only if there exists $n \in \natp$ such that $f_{n}(x) > \alpha$. Thus
(2): Let $\alpha \in \real$, then
By Proposition 17.2.4 and Lemma 21.1.5, $F$ and $f$ are both $(\cm, \cb_{\ol \real})$-measurable.
(3): $\limsup_{n \to \infty}f_{n} = \inf_{n \in \natp}\sup_{k \ge n}f_{k}$.
(4): $\liminf_{n \to \infty}f_{n} = \sup_{n \in \natp}\inf_{k \ge n}f_{k}$.
(5): If $\limv{n}f_{n}$ exists, then it is equal to (3) and (4). In which case, it is measurable.
Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_{\real})$-measurable by Lemma 21.3.1.$\square$
Lemma 21.3.3.label Let $f: \real \to \real$ be a non-decreasing or non-increasing function, then $f$ is Borel measurable.
Proof. By taking $-f$, assume without loss of generality that $f$ is non-decreasing. In which case, for any $a \in \real$, $x \in f^{-1}((-\infty, a])$, and $y \le x$, $f(y) \le f(x) \le a$, so $y \in f^{-1}((-\infty, a])$. Thus $f^{-1}((-\infty, a))$ is an interval and hence measurable. Since the open rays generate $\cb_{\real}$ (Proposition 17.2.3), $f$ is Borel measurable.$\square$