I'm trying to get a good handle on analysis counterexamples as they relate to convergence in $M(X)$ and $C_0(X)$. Awhile back there was an excellent discussion of pointwise convergence, convergence in $L^p$ norm, weak convergence in $L^p$ and convergence in measure, here.

How about a similar set of counterexamples for $M(X)$ and $C_0(X)$? Here we have the notions of vague convergence, weak* convergence, convergence in norm (where the norm of a complex Radon measure is its total variation).

How about these counterexamples (asked in Folland or a variation therein):

a) $\mu_n\to 0$ vaguely, but $\|\mu_n\|=|\mu|(X)|\nrightarrow 0$.

b) $\mu_n\to 0$ vaguely, but $\int f\ d\mu_n\nrightarrow \int f\ d\mu$ for some bounded measurable $f$ with compact support.

c) $\mu_n\ge 0$ and $\mu_n\to 0$ vaguely, but $\mu_n((-\infty,x])\nrightarrow \mu((-\infty,x])$ for some $x\in\mathbb{R}$.

d) $\{f_n\}\in C_0(X)$ converges weakly to some $f$, but not pointwise.

Any links to conceptual ways of internalizing these different notions of convergence, in addition to providing counterexamples, would be greatly appreciated.


EDIT: Let me define the notions of convergence as Folland does:

Vague convergence means convergence with respect to the vague topology on $M(X)$, which is also known as the weak* topology on $M(X)$, which means that $\mu_\alpha\to \mu$ iff $\int f\ d\mu_\alpha\to \int f\ d\mu$ for all $f\in C_0(X)$.

Weak convergence means that convergence on $X$ with respect to the topology generated by $X^*$.

The norm on $M(X)$ is given by the total variation, so that $\|\mu\|=|\mu|(X)$, so that convergence with respect to this norm means that $|\mu_n-\mu|(X)\to 0$.

Incidentally I find it awkward working with the total variation in such discussions of convergence because there is no clear geometry to work with as far as I can tell; the definition is rather too abstract for me at the moment.

  • $\begingroup$ Can you give your definitions for the various modes of convergence? Authors sometimes vary in their usage. Anyway, as a hint, consider $\delta_{x_n}$, a measure with a unit point mass at some $x_n \in \mathbb{R}$, and consider what happens as $x_n \to x$ or $x \to \pm \infty$. $\endgroup$ – Nate Eldredge Mar 25 '12 at 18:10
  • $\begingroup$ Thanks, Nate. I added the definitions to the question - I am using Folland as my guide through analysis. I actually considered that point mass measure, possibly from some other resource, but I can't find a way to make it work with the definitions. $\endgroup$ – user21725 Mar 25 '12 at 20:01
  • $\begingroup$ So let's start with part (a). Let $\mu_n = \delta_n$ be a point mass at $n$. I claim that this sequence has the desired properties. Can you verify this from the definitions? If not, where do you get stuck? $\endgroup$ – Nate Eldredge Mar 25 '12 at 23:46
  • $\begingroup$ The total variation of a complex Radon measure is the positive measure $|\nu|$, where $d|\nu|=|f| d\mu$ for $\mu$ a positive measure, where by the Riesz Representation theorem $f$ is in $C_0(X)$. So if $\mu_n=\delta_n\to 0$ as $n\to\infty$, s.t. $|\mu_\alpha|(X)\to |\mu|(X)$ this means that for some $f\in C_0(X)$, $f(n)=|\mu_\alpha|=\int |f_\alpha|\ d\mu \nrightarrow \int |f|\ d\mu=|\mu|(X)=0$, but $\delta_n\to 0$ vaguely since for all $f\in C_0(X)$ for $n$ large $\int f\mu_n$ is small. How is that? $\endgroup$ – user21725 Mar 26 '12 at 2:59
  • $\begingroup$ Your definition of total variation seems a little confused, you might want to review the definition. However it is worth noting that for a positive measure $|\mu| = \mu$. So in fact we have $\lVert \delta_n \rVert = 1$ for all $n$. It is also worth knowing that for a complex Radon measure $\mu$, $\lVert \mu \rVert = \sup\{\int f \,d\mu : f \in C_0(X), \lVert f \rVert_\infty \le 1\}$, i.e. the total variation norm is the operator norm in $C_0(X)^*$. $\endgroup$ – Nate Eldredge Mar 26 '12 at 3:07

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