Proof. Let $\seqi{f}\subset L^{\infty}(X; \real)$ and $M \in \real$ such that $f_{i} \le M$ almost everywhere for all $i \in I$.

Fix $A \in \cm$ with $\mu(A) < \infty$, and let

then since $f_{i} \le M$ almost everywhere for all $i \in I$, $\mathcal{S}_{A} \ne \emptyset$, and $m_{A} = \inf_{g \in \mathcal{S}_A}\int g d\mu \in \real$.

Let $\seq{g_{A, n}}\subset \mathcal{S}_{A}$ such that $\seq{g_{A, n}}$ is decreasing pointwise and $\limv{n}\int_{A} g_{A, n}d\mu \downto m_{A}$. Take $g_{A} = \limv{n}g_{A, n}$, then by the Dominated Convergence Theorem, $\int g_{A} d\mu = m_{A}$.

For each $i \in I$, since $g_{A, n}\ge f_{i}|_{A}$ almost everywhere for all $n \in \natp$, $g_{A} \ge f_{i}|_{A}$ almost everywhere as well. Thus $g_{A} \in \mathcal{S}_{A}$. For any $h \in \mathcal{S}_{A}$, $g_{A} \wedge h \in \mathcal{S}_{A}$ with

Thus $g_{A} \wedge h = g_{A}$ almost everywhere, so $g_{A} \le h$ almost everywhere, and $g_{A}$ is an essential supremum of $\bracsn{f_i|_A}_{i \in I}$.

Now, let $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$, then $\one_{A \cap B}g_{B} + \one_{A \setminus B}M \in \mathcal{S}_{A}$, and

Thus $g_{A} \wedge (\one_{A \cap B}g_{B} + \one_{A \setminus B}M) = g_{A}$ almost everywhere, so $g_{A}|_{A \cap B}\le g_{B}|_{A \cap B}$ almost everywhere. As the argument is symmetric, $g_{A}|_{A \cap B}= g_{B}|_{A \cap B}$ almost everywhere.

By the gluing lemma for measurable functions, there exists a measurable function $g:X \to \real$ such that $g|_{A} = g_{A}$ for all $A \in \cm$ with $\mu(A) < \infty$.

Let $h \in L^{\infty}(X; \real)$ with $h \ge f_{i}$ almost everywhere for all $i \in I$, then for any $A \in \cm$ with $\mu(A) < \infty$,

As $\mu$ is semifinite, $\mu(\bracs{h < g}) = 0$. Finally, since $g_{A} \le M$ almost everywhere for all $A \in \cm$ with $\mu(A) < \infty$, $g \le M$ almost everywhere. Therefore $g \in L^{\infty}(X; \real)$ is indeed the essential supremum of $\seqi{f}$.$\square$