# proof of existence of the essential supremum

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

$$\mathrm{ess}sup\mathcal{S}=\underset{n}{sup}{f}_{n}.$$ |

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 $\mathcal{S}$ is nonempty, and let ${\mathcal{S}}^{\prime}$ consist of the collection of maximums of finite sequences of functions in $\mathcal{S}$. Then choose any continuous^{} and strictly increasing $\theta :\overline{\mathbb{R}}\to \mathbb{R}$. For example, we can take

$$ |

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

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

Then choose a sequence ${g}_{n}$ in ${\mathcal{S}}^{\prime}$ such that $\int \theta ({g}_{n})\mathit{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 $\mathcal{S}$. Therefore, there exists a sequence ${f}_{n}\in \mathcal{S}$ 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 $\mathcal{S}$. First, by continuity of $\theta $ and the dominated convergence theorem,

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

Similarly, for any $g\in \mathcal{S}$,

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

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 $g:\mathrm{\Omega}\to \overline{\mathbb{R}}$ satisfies $g\ge h$ ($\mu $-a.e.) for all $h\in \mathcal{S}$. Then, $g\ge {f}_{n}$ and,

$$g\ge \underset{n}{sup}{f}_{n}=f$$ |

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

Title | proof of existence of the essential supremum |
---|---|

Canonical name | ProofOfExistenceOfTheEssentialSupremum |

Date of creation | 2013-03-22 18:39:25 |

Last modified on | 2013-03-22 18:39:25 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 7 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 28A20 |