Pointwise Supremum of Measurable Functions is Measurable
Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space, and let $I$ be a countable set.
Let $\left({f_i}\right)_{i \in I}$, $f_i: X \to \overline \R$ be an $I$-indexed collection of $\Sigma$-measurable functions.
Then the pointwise supremum $\displaystyle \sup_{i \mathop \in I} f_i: X \to \overline \R$ is also $\Sigma$-measurable.
Proof
Let $a \in \R$; for all $i \in I$, we have by Characterization of Measurable Functions that:
- $\left\{{f_i > a}\right\} \in \Sigma$
and as $\Sigma$ is a $\sigma$-algebra and $I$ is countable, also:
- $\displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \in \Sigma$
We will now show that:
- $\displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\} = \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\}$
First, observe that for all $i \in I$:
- $f_i \left({x}\right) \le \displaystyle \sup_{i \mathop \in I} f_i \left({x}\right)$
and hence:
- $\left\{{f_i > a}\right\} \subseteq \displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$
From Union is Smallest Superset: Family of Sets:
- $\displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \subseteq \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$
Next, suppose that:
- $x \notin \displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\}$
Then, by definition of union:
- $\forall i \in I: f_i \left({x}\right) \le a$
which is to say that $a$ is an upper bound for the $f_i \left({x}\right)$.
Hence, by definition of supremum, it follows that:
- $\displaystyle \sup_{i \mathop \in I} f_i \left({x}\right) \le a$
and therefore:
- $x \notin \displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$
Thus, we have shown:
- $\displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\} = \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \in \Sigma$
and by Characterization of Measurable Functions, it follows that $\displaystyle \sup_{i \mathop \in I} f_i$ is measurable.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $8.9$