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:


 * $\ds \set {\sup_{i \mathop \in I} f_i > a} = \bigcup_{i \mathop \in I} \set {f_i > a} \in \Sigma$

and by Characterization of Measurable Functions, it follows that $\ds \sup_{i \mathop \in I} f_i$ is measurable.