$\renewcommand{\bar}{\overline}
\DeclareMathOperator{\Lim}{Lim}
\newcommand{\N}{\mathbf{N}}
\newcommand{\R}{\mathbf{R}}$

Hello dearest readers! How a joyful day it was, full of good math. Actually it sucked. In any case, I have tried to post some of the things I have written on my blog. This is my first attempt on pushing the document which gives a finitely, but not countable additive measure. It will require more explanation, but yours truly is truly unsatisfied with the typesetting and needs to start some decent scripting before this gets done. Hold on to your nickers, this will be updated soon.

Why then do I post it you say? Good question! I will attempt to answer it. The main reason is that it pushes me harder out of inertia, and hence improving myself. I like that, pushing my own boundaries such that they are no boundaries anymore and are then elsewhere. The other thing is that some might have some suggestions on how to improve the LaTeX -> WordPress thing.

Edit: The issues have been resolved! I now have my own WordPress setup, so I can use MathJax. I set up a different VPS for that as I do not want to poison my other servers with filthy things like PHP. Got it? Good! In any case, it looks better now. So much better, I removed the ‘try #1′ from the title. I even have footnotes! How cool is that?

A finitely but not countably additive measure

Consider the space $\ell^\infty$ of all bounded real-valued sequences on the natural numbers $\N = \{1, 2, 3, \dots\}$. Let $c$ be the subspace of $\ell^\infty$ of all convergent sequences. Next, we define the operator $\Lim$ on $c$, which is given by:
\begin{equation}
\label{eq:Lim-operator}
\begin{split}
\Lim : c &\to \R\\
x &\mapsto \lim_{n \to \infty} x_n.
\end{split}
\end{equation}
For the purpose of extending $\Lim$ to $\ell^\infty$, we define the mapping $p$:
\begin{equation}
\label{eq:Extension-of-Lim-Hahn-Banach}
p(x) := \inf\Bigl\{\Lim(y) : y \in c \, \text{and} \,
\limsup_{n \to \infty} x_n \leq \lim_{n \to \infty} y_n \Bigr\}.
\end{equation}
Note that $p$ is defined for all in $\ell^\infty$ as certainly we can find for every bounded real-valued sequence a convergent sequence whose limit is above any term in the sequence. The subadditivity $p(x + y) \leq p(x) + p(y)$ is easily established. Whenever $x$ is in $c$, then we have that $p(x) = \Lim(x)$, hence we can apply the Hahn-Banach theorem. By this theorem there exists a linear extension of $\Lim$ which we will henceforth denote by $\Lim$ as well, or $\overline\Lim$ when confusion might arise. Additionally, there holds that:
\begin{equation*}
\overline\Lim(x) \leq p(x) \quad \text{for all $x$ in $\ell^\infty$.}
\end{equation*}
Setting $x \to -x$ gives $\overline\Lim(-x) \leq p(-x)$. Hence:
\begin{equation*}
-p(-x) \leq \overline\Lim(x) \leq p(x).
\end{equation*}
Whenever $x$ is in $c$, then ${\overline\Lim}(x) = \Lim(x)$, hence we consider $x$ in $\ell^\infty \setminus c$. Using \eqref{eq:Extension-of-Lim-Hahn-Banach} and $x \geq 0$ we get:
\begin{align*}
-p(-x) &= -\inf\Bigl\{-\Lim(-y) : y \in c \, \text{and} \, \limsup_{n \to \infty}
-x_n \leq \lim_{n \to \infty} y_n \Bigr\}\\
&= \sup\Bigl\{\Lim(-y) : y \in c \, \text{and} \, \limsup_{n \to \infty}
-x_n \leq \lim_{n \to \infty} y_n \Bigr\}.
\end{align*}
In the last line we can set $y = 0$ in the supremum to obtain a lower bound, and get:
\begin{equation*}
0 \leq -p(-x) \leq \overline\Lim(x).
\end{equation*}
In particular if $x \geq 0$, then $\Lim(x) \geq 0$, so $\Lim$ is a positive operator defined on all of $\ell^\infty$. Furthermore, $\Lim$ is a so-called Banach limit.

Next, we consider the $\sigma$-algebra on $\N$ which consists of all its subsets, that is $\sigma(\N) = 2^\N$. We now define a finitely additive measure $\mu : \sigma(\N) \to \{0, 1\}$ for every subset $S$ of $\N$ as follows¹:
\begin{equation*}
\mu(S) = \overline\Lim \bigl([n \in S] \bigr).
\end{equation*}
We can now see that $\mu$ is a finitely additive measure. $\mu(\emptyset) = 0$ and $\mu \geq 0$ are easy to verify. For the finite additivity select a finite collection of disjoint sets $\{S_n\}_{n = 1}^k$. These $S_n$ are finite or infinite sequences. So, by linearity of $\Lim$ we get:
\begin{equation*}
\mu\biggl(\bigcup_{n = 1}^k S_n \biggr) = \Lim\biggl(\sum_{n = 1}^k
S_n \biggr) = \sum_{n = 1}^k \Lim(S_n) = 0.
\end{equation*}
To show $\mu$ is not $\sigma$-additive, consider the sets $S_n := \{n\}$. The singletons containing the natural number $n$. As these are finite sequences, their limit is $0$. That is:
\begin{equation*}
\sum_{n = 0}^\infty \mu(S_n) = \sum_{m = 0}^\infty
\Lim\bigl([m = n]\bigr) = 0.
\end{equation*}
While we have for their union:
\begin{equation*}
\mu\biggl(\bigcup_{n \in \N} S_n \biggr) = \mu\bigl([n \in \N]\bigr)
= 1.
\end{equation*}
Hence, $\mu$ is not countable additive.

Kisses, Cheers, So long and thanks for all the fish,

JT.

A pdf version of this document can be obtained here: Finitely additive measure.