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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.

Thanks.

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.

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    Folland defines the total variation of $\nu$ (page 93) as the unique positive measure $|\nu|$ such that if $\mu$ is a positive measure, $f$ is a measurable function, and $d\nu = f d\mu$, then $d|\nu| = |f| d\mu$. I don't see why you are bringing in the Riesz representation theorem, and I don't understand what $\mu_\alpha$ and $\mu$ are supposed to be. Unfortunately this is not a great forum for working through details like this.2012-03-26

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