I am trying to solve this question:
Give an example of a measure space $(X, \mathfrak{M}, \mu)$ for which the Riesz Representation Theorem does extend to the case $p=\infty.$
My Trial:
I am trying to mimic the proof in Royden "Real Analysis" fourth edition on page 402.
My example is the space of 1 point i.e. $X = \{x_{0}\}$ with $\mu$ the counting measure. (taking into account that $L^{\infty}$ in this case is the space of all bounded measurable functions $f: X \rightarrow \mathbb{R}$ which is the space of constant functions, and that $L^1$ in this case is the collection of integrable functions of $X$ but we know that $X$ is the space of all constant functions which are all integrable.)
Proof:
Since our counting measure equals 1, so we are in the case $\mu(X)< \infty.$ Let $S:L^{\infty}(X, \mu) \rightarrow \mathbb{R}$ be a bounded linear functional. Define a set function $\nu$ on the collection of measurable sets $\mathfrak{m}$ by setting $$\nu(E) = S(\chi_{E})$$ for $E \in \mathfrak{M}.$this is properly defined since $\mu(X) < \infty$ and thus the characteristic function of each measurable set belongs to $L^{\infty}(X, \mu).$ We claim that $\nu$ is a signed measure.i.e. we need to show that $$\nu(E) = \sum_{k=1}^{\infty} \nu(E_{k})$$ and the series converges absolutely.Indeed, let $\{E_{k}\}_{k=1}^{\infty}$ be a countable disjoint collection of measurable sets and $E = \bigcup_{k=1}^{\infty}E_{k}.$ By the countable additivity of the measure $\mu,$ $$\mu(E) = \sum_{k=1}^{\infty} \mu(E_{k}) < \infty.$$ therefore $$\lim_{n\rightarrow \infty} \sum_{k= n+1}^{\infty} \mu(E_{k}) = 0.$$
Consequently, $|\nu(E) - \sum_{k=1}^{n}\nu(E_{k})| = |S(\chi_{E}) - \sum_{k=1}^{n}S(\chi_{E_{k}})| = |S(\chi_{E} - \sum_{k=1}^{n} \chi_{E_{k}})| $
.....
I am just going to mimic the proof in the book.
My question is:
Is there is something special about my example that makes the proof easier (as in the book the proof contains finding the function $f\in L^1$as the radon Nikodym derivative of $\nu$ with respect to $\mu$)