proof of existence of the essential supremum

Suppose that $(\mathrm{\Omega },ℱ,\mu )$ is a $\sigma$-finite measure space and $𝒮$ is a collection of measurable functions $f:\mathrm{\Omega }\to \overline{ℝ}$. We show that the essential supremum of $𝒮$ exists and furthermore, if it is nonempty then there is a sequence $f_n\in 𝒮$ such that

$$\mathrm{ess}sup𝒮=\underset{n}{sup}{f}_n.$$

As any $\sigma$-finite measure is equivalent to a probability measure (http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure), we may suppose without loss of generality that $\mu$ is a probability measure. Also, without loss of generality, suppose that $𝒮$ is nonempty, and let ${𝒮}^{\prime }$ consist of the collection of maximums of finite sequences of functions in $𝒮$. Then choose any continuous and strictly increasing $\theta :\overline{ℝ}\to ℝ$. For example, we can take

As $\theta (f)$ is a bounded and measurable function for all $f\in {𝒮}^{\prime }$, we can set

$$\alpha =sup\{\int \theta (f)𝑑\mu :f\in {𝒮}^{\prime }\}.$$

Then choose a sequence $g_n$ in ${𝒮}^{\prime }$ such that $\int \theta (g_n)𝑑\mu \to \alpha$. By replacing $g_n$ by the maximum of $g_1,\mathrm{\dots },g_n$ if necessary, we may assume that $g_{n+1}\ge g_n$ for each $n$. Set

$$f=\underset{n}{sup}{g}_n.$$

Also, every $g_n$ is the maximum of a finite sequence of functions $g_{n,1},\mathrm{\dots },g_{n,m_n}$ in $𝒮$. Therefore, there exists a sequence $f_n\in 𝒮$ such that

$$\{f_1,f_2,\mathrm{\dots }\}=\{g_{n,m}:n\ge 1,1\le m\le m_n\}.$$

Then,

$$f=\underset{n}{sup}{f}_n.$$

It only remains to be shown that $f$ is indeed the essential supremum of $𝒮$.First, by continuity of $\theta$ and the dominated convergence theorem,

$$\int \theta (f)𝑑\mu =\underset{n\to \mathrm{\infty }}{lim}\int \theta (g_n)𝑑\mu =\alpha .$$

Similarly, for any $g\in 𝒮$,

$$\int \theta (f\vee g)=\underset{n\to \mathrm{\infty }}{lim}\int \theta (g_n\vee g)𝑑\mu \le \alpha .$$

It follows that $\theta (f\vee g)-\theta (f)$ is a nonnegative function with nonpositive integral, and so is equal to zero $\mu$-almost everywhere. As $\theta$ is strictly increasing, $f\vee g=f$ and therefore $f\ge g$ $\mu$-almost everywhere.

Finally, suppose that $g:\mathrm{\Omega }\to \overline{ℝ}$ satisfies $g\ge h$ ($\mu$-a.e.) for all $h\in 𝒮$. Then, $g\ge f_n$ and,

$$g\ge \underset{n}{sup}{f}_n=f$$

$\mu$-a.e., as required.