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.)


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 $\mu$)

  • 1
    $\begingroup$ Yes, this vector space is finite (one) dimensional, and in a finite dimensional vector space any two norms are topologically equivalent. $\endgroup$ – Berci Feb 17 at 8:26
  • $\begingroup$ So what? how can I complete the proof? @Berci I need to proof the Onto part, I need an isomorphism. what are the amendments that I should make to the given proof in the book? $\endgroup$ – Mathstupid Feb 17 at 10:04
  • $\begingroup$ It is Riesz representation theorem for the dual of $L^p(X,\mu)$@Berci $\endgroup$ – Mathstupid Feb 17 at 10:08
  • $\begingroup$ @Berci how about the function $f \in L^1$ that I should construct? $\endgroup$ – Mathstupid Feb 17 at 10:21
  • $\begingroup$ @Berci Do we have to mimic the proof of the Riesz Representation theorem for the dual of $L^p({x_{0}}, \mu)$ in case of Lebesgue measure instead (on pg.160 of the mentioned book)? $\endgroup$ – Secretly Feb 17 at 16:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.