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