Talk:Convergence of measures

Texxed. Introduced the notion of total variation measure to simplify notation and definitions. Corrected a mistake in the definition of total variation (the original one worked only for real-valued measures). Added the Riesz representation theorem and connection to Banach space theory. Precised some terminology. I have a real analysis background in measure theory: I kept the original terminology, but also added the one which is more common in real analysis. Some real analysts love to quarrel with probabilists about terminology and how measure theory should be taught: I am not one of them and open to a friendly debate with fellows probabilists :-). Camillo 13:49, 21 July 2012 (CEST)

I do not take the hint... maybe because I am both a probabilist and a (real?) analyst... do I quarrel with myself? I only know that indeed probabilists say "weak" when functional analysts say "weak$^\star$". Yes, there is a clash here. But let me ask, is the weak (no star) topology on measures really useful?

Another question. On the set of probability measures some (or all?) of these topologies (except for the strong one, of course) should coincide. (For now I am afraid to say which exactly.)

Boris Tsirelson 19:27, 21 July 2012 (CEST)

Hello, welcome (and if you don't want to remain anonymous, just add four tildas at the end of your message and you get your signature automatically). After all I am not really a real analyst, but a PDE guy: this maybe answers to your first question :-). Yes, I think the weak topology is utterly useless... I just kept it because the original article had it. Even for probability measures, I think all topologies are indeed different, because:

* If $x_n$ is a sequence of real numbers going to infinity, then in the topology 4) the sequence $\delta_{x_n}$ converges to $0$, but this is not true in the topology 3). Of course 3) and 4) do coincide when the underlying space is compact, but then this is true for general measures (as already remarked).

* As for the difference between the topologies 2) and 3), consider again $X=\mathbb R$. The functional $\mu\mapsto L (\mu) := \mu (]0,1[)$ is linear and continuous. So, the sequence $\delta_{1/n}$ converges in the topologies 3) and 4) to the $\delta_0$, but it does not converge in the topology 2). Camillo 18:30, 8 August 2012 (CEST)

Sorry, I forgot to sign... now I did. Yes, I see, (2) and (3) differ. But for (3) and (4) I am not convinced, since 0 is not a probability measure. Think about two topologies on [0,1]: the usual one, and another one that makes 0 an isolated point. They differ on [0,1] but still coincide on (0,1]. --Boris Tsirelson 08:16, 9 August 2012 (CEST)

Yes, now I recall the argument; (3) and (4) coincide on probability measures. That is, every narrow neighborhood of a probability measure $\mu$ is also a wide neighborhood of $\mu$. Here is a sketch of a proof. Given $\varepsilon$, we take a compactly supported continuous $f:X\to[0,1]$ such that $\int f \rd\mu > 1-\varepsilon$. Now, consider another probability measure $\nu$ widely close to $\mu$, namely, satisfying $|\int f \rd(\mu-\nu)| < \varepsilon$ and therefore $\int f \rd\nu > 1-2\varepsilon$.

Claim: If such $\nu$ satisfies $|\int fg \rd(\nu-\mu)|<\varepsilon$ for a given $g:X\to[-1,1]$ then $|\int g \rd(\nu-\mu)|<4\varepsilon$.

Proof of the claim: $ |\int g \rd(\nu-\mu) - \int fg \rd(\nu-\mu)| = |\int (1-f)g \rd(\nu-\mu)| \le |\int (1-f)g \rd\nu| + |\int (1-f)g \rd\mu| \le \int (1-f) \rd\nu + \int (1-f) \rd\mu < 2\varepsilon + \varepsilon $.

--Boris Tsirelson 10:47, 9 August 2012 (CEST)

Indeed I also recalled your argument above: the convergence of the mass prevents escape to infinity, but I was typing my answer while you wrote the last comment (and my example just shows that the closures of the space of probability measures differ in the largest space when we change the two topologies). I will add a remark. I also noticed that a topological assumption is missing for the remark on (3).Camillo 11:02, 9 August 2012 (CEST)

Hi Camillo -- a couple of suggestions:

• Should $\mu: \mathcal{B}\mapsto \mathbb R$ actually be $\mu: \mathcal{B}\rightarrow \mathbb R$?
• Perhaps replace $\mathcal{B}$ by $\mathscr{B}$?

--Jjg 14:35, 21 July 2012 (CEST)

Hi Jjg. I agree with the first one (just done). The second is a matter of typographical taste: I have nothing against it, but I am not keen in chasing all the $\mathcal{B}$'s to replace them... in other words, you would be welcome to do it :-)) Camillo 14:40, 21 July 2012 (CEST)