In Measure and Integration Theory class we saw the following definition of an essential supremum of a collection of functions:

Definition: Let $(f_{\omega})_{\omega \in \Omega}$ be a collection of measurable functions mapping from $X$ to $\overline{\mathbb{R}}$. Then a measurable map $h$ is called the essential supremum of $(f_{\omega})_{\omega \in \Omega}$, $$h=esssup_{\omega \in \Omega}f_{\omega}$$, if it satisfies $f_{\omega} \leq h$ $\mu$-a.e. for all $\omega \in \Omega$ and if $h \leq h'$ $\mu$-a.e. for every measurable $h'$ from $X$ to $\overline{\mathbb{R}}$ which satisfies $f_{\omega} \leq h'$ $\mu$-a.e. for all $\omega \in \Omega$. Where $\Omega$ is an arbitrary Index set.

Then we were given the following theorem:

Theorem: Suppose that $\mu$ is $\sigma$-finite. For every family $(f_{\omega})_{\omega \in \Omega}$ the essential supremum $esssup(f_{\omega})_{\omega \in \Omega}$ exists and is $\mu$-a.e. unique. Moreover, there exists a countable set $J \subseteq \Omega$ with $esssup(f_{\omega})_{\omega \in \Omega}=\sup_{j \in J}f_{i}$ pointwise.

My confusion was with the following part of the proof. We defined $\mathcal{A}:=\{ A \subseteq \Omega : A$ Is countable $\}$. Then for each $A \in \mathcal{A}$, set $g_{A} := \sup_{j \in A} f_{j}$, $c_{A} := \int g_{A} d \mu$ and $c:=\sup_{A \in \mathcal{A}} c_{A}$. Let $(A_{n})_{n \in \mathbb{N}}$ be a sequence in $\mathcal{A}$ with $c_{A_{n}} \nearrow c$.......

1) How does such a sequence of sets $(A_{n})_{n \in \mathbb{N}}$ even exist if the set $\mathcal{A}$ does not have to be countable nor ordered?

2) Or is there maybe another source on how to prove this? I searched through books in our Library at uni but could not find even the definition in the Measure and Integration books.

Thanks a lot in advance!

  • axiom of choice? – Euler....IS_ALIVE May 2 '17 at 16:08
  • To my understanding, the existence of a Choice Function on the set $\mathcal{A}$ (hence using the axiom of choice) does not imply that one chooses the sequence to be strictly increasing nor converging to to $c$. But hey I could be wrong I'm just in my second year of studying mathematics. – vaoy May 2 '17 at 16:18
  • Every bounded sequence of numbers contains a convergent subsequence to the supremum (you can just use the definition to prove this). Now just take the corresponding sets from $\mathcal{A}$ – Euler....IS_ALIVE May 2 '17 at 17:18
  • Hmm... I still don't see how I could construct such a sequence of the sets. Could you maybe explain a little further? What bounded sequence of numbers are you referring to? – vaoy May 2 '17 at 18:31
  • @Euler....IS_ALIVE oooooh I see it now. Thank you very much! – vaoy May 4 '17 at 7:12
up vote 1 down vote accepted

As it was pointet out by @Euler....IS_ALIVE, in the comments to my question, one needs to use the Axiom of choice or the weaker form of it, the Axiom of countable choice. Let me explain, because one can use the following Theorem:

One can construct a set $\{c_{A}:A \in \mathcal{A}\}$ set of $c_{J}$'s. In the proof $c_{A} := \int g_{A} d \mu$ is bounded by $\mu(X)< \infty$ from above for all $A \in \mathcal{A}$, hence for every countable subset of the index set $\Omega$.

Now one can apply the following Theorem

Theorem: Every subset $M \subset \mathbb{R}$ which is bounded from above by a real number $a \in \mathbb{R}$ has a convergent subsequence to $\sup{M}$, where $\sup$ denotes the supremum.

The proof idea goes as follows:

If $\sup{M} \in M$ then just set $x_{n}=\sup{M}$ for all $n \in \mathbb{N}$.

If $\sup{M}$ is not in $M$ then fix an Element $x_{0}$ in $M$ assuming that $M$ is not empty. Then by Definition of the supremum there exists an Element in $\Lambda_{1} = \{ x \in M : x> \sup{M} - x_{0} \}$. Using the axiom of choice (or the weaker version "Axiom of countable choice") one can choose an Element $x_{1} \in \Lambda_{1}$. Since by Definition of the supremum such an Element must exists. Now one can continue this and for a general $n \in \mathbb{N}$ there exists an $x_{n} \in \Lambda_{n}=\{x \in M: x> \sup{M} - x_{n-1}\}$. Therefore one has constructed an increasing sequence $x_{n} \leq x_{n+1}$ that converges to $\sup{M}$.

This theorem can now be used to assert existence of such a countable sequence $c_{A_{n}}$ in $\{c_{A}:A \in \mathcal{A}\}$ which converges to $c$. Then the set of index sets $(A_{n})_{n \in \mathbb{N}}$ also exists and every $A_{n}$ is countable.

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