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This is probably a very easy question for experts in probability or measure theory.

I have a sequence of finite measures $\mu_{n}$ on a non-compact metric space $X$ such that $\mu_{n}$ converges to $\mu$ in the following sense: $$ \int_{X}fd\mu_{n} \to \int_{X}fd\mu \ \ \ \ \ \text{ for all f continuous with compact support} $$ I would like to say that $\mu_{n}(X)\to \mu(X)$.

I know this is false in general, but I have the additional condition that for every $\epsilon>0$ there is $n_{0}\in \mathbb{N}$ and $K\subset X$ compact such that $\mu_{n}(K^{c})\leq \epsilon$ for every $n\geq n_{0}$. This looks very similar to the definition of tight sequence (which guarantees the result I would like). Is this equivalent?

Additional assumptions: X is Polish and locally compact, precisely it is a closed surface with some finitely many points removed. All measures $\mu_{n}$ and $\mu$ are area measures of Riemannian metrics (with singularities at the points removed) on X.

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  • $\begingroup$ I would say yes, but I'm not sure what the "official" definition of tightness is you are using. $\endgroup$ Jun 16, 2020 at 16:23
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    $\begingroup$ The definition of tightness I found is the following: a sequence $\mu_{n}$ is tight if for every $\epsilon>0$ there is a compact set $K$ such that $\mu_{n}(K^c)<\epsilon$ for all $n$. The difference with what I have is that for me this condition holds only for $n \geq n_{0}$ with $n_{0}$ depending on $\epsilon$. $\endgroup$
    – AMath91
    Jun 16, 2020 at 16:26
  • $\begingroup$ Is your metric space by any chance separable and complete (or each of your measures regular)? Then you can find for each $m<n_0$ a compact set $K_m$ such that $\mu_m(K_m^c)\leq \epsilon$. Replacing $K$ with $K\cup K_1\cup\cdots\cup K_{n_0-1}$ would show that the sequence is tight in the usual sense. $\endgroup$ Jun 16, 2020 at 16:39
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    $\begingroup$ Say the metric space is $\mathbb Q$ with the usual topology. It has no nonzero continuous functions with compact support. So $$\int_{X}fd\mu_{n} \to \int_{X}fd\mu \quad\text{ for all f continuous with compact support}$$is vacuously true, but $\mu_{n}(X)\to \mu(X)$ could easilyt fail. $\endgroup$ Jun 16, 2020 at 16:53
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    $\begingroup$ Even $X$ being Polish, which is pretty much as nice as it gets, doesn't guarantee that you have non-zero continuous functions with compact support, think of $X = \ell^2$. You probably want $X$ to be Polish and locally compact, and the $\mu_n$'s to be positive, but then the claim is of course trivial. $\endgroup$ Jun 16, 2020 at 18:29

2 Answers 2

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Since your space is Polish, $\mu$ is regular and there exists a compact set $C$ such that $\mu(X\setminus C)<\epsilon$ for each $\epsilon>0$. Since your space is locally compact, there is another compact set $C'$ such that $C$ is a subset of the interior of $C'$. The function given by $f_n(x)=\max\{0,1-n\cdot d(C,x)\}$ is continuous and supported on $C'$ whenever $n>d(C,C')^{-1}$. Moreover, the sequence $\langle f_n\rangle$ decreases pointwise to the indicator function$1_C$. It follows that $\liminf_n \mu_n(X)\geq\mu(X)$ for each $\epsilon>0$ and $n$ large enough. A similar argument applied to the compact set in your tightness version shows that $\limsup_n\mu_n(X)\leq\mu(X)$.

That the sequence is tight follows directly from inner regularity of measures on Polish spaces. If there $n_{0}\in \mathbb{N}$ and $K\subset X$ compact such that $\mu_{n}(X\setminus K)\leq \epsilon$ for every $n\geq n_{0}$, you can just pick a compact set $K_m$ for each $m< n_0$ such that $\mu_m(X\setminus K_m)\leq\epsilon$. Take $K'=K\cup\bigcup_{m<n_0}K_m$, then $\mu_n(X\setminus K')\leq\epsilon$ for all $n$. As has been pointed out in the comments, tightness is not enough if you only have convergence of integrals for compactly supported continuous functions.

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This is really just a comment to add context to your question and the reactions but it will be too long so I am disguising it as an answer. Stripping away all trappings the relevant fact is that an equicontinuous set $A$ in the dual of a locally convex space $E$ is relatively compact for the corresponding weak topology and so that latter coincides on $A$ with the weak topology induced by any dense subspace $E_0$ of $E$. In your case, $E$ is the space of bounded, continuous functions on a locally compact space, provided with the so-called strict topology which was introduced by R.C. Buck in the 50‘s. This has the following three relevant properties which set up the connection to your question.

The dual space is the space of finite tight (or Radon) measures;

A family of measures is equicontinuous if and only if it is bounded and uniformly tight;

The space $E_0$ of continuous functions with compact supports is dense.

This shows that many of the assumptions given above are irrelevant. One can even replace local compactness by tcomplete regularity. The former condition is required to ensure the denseness of $E_0$. In the general case, one can use any subalgebra of $E$ which separates points and is such that for each element of the underlying space there is a function in $E_0$ which does not vanish there.

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  • $\begingroup$ The original formulation did not come with the assumption that the measures are Radon. $\endgroup$ Jun 18, 2020 at 18:36
  • $\begingroup$ The formulation in the question is imprecise—it wasn’t clear whether the measure are probability measures (presumably not), positive, or signed (in which case there is an absolute value sign missing in the definition of uniform tightness). This means that there is some ambiguity in the statement. It seemed reasonable to me to assume that since the OP was using the phrase tightness for sequences of measure, that he was tacitly assuming that the individual measures were tight. If not, what are they? Maybe he could specify what regularity condition he understands in the term “measure”. $\endgroup$
    – user131781
    Jun 18, 2020 at 18:48
  • $\begingroup$ The point of asking whether it is a Polish space is exactly to ensure they are Radon. The test functions used are not bounded continuous functions but compactly supported functions, which is why local compactness matters. You write that "many of the assumption given above are irrelevant" but simply replace them by others. $\endgroup$ Jun 18, 2020 at 18:57
  • $\begingroup$ No. As I tried to explain, the original formulation is not precise enough to specify what concept of measure is intended. Is it finite p, $\sigma$- or $\tau$-additivity, tightness, or ..... I chose what seemed the most natural interpretation but I could, of course, be wrong. Only the OP can know—perhaps he could put us out of our misery by editing his post. $\endgroup$
    – user131781
    Jun 18, 2020 at 19:06
  • $\begingroup$ They are finite measures on a Polish locally compact space; as is written in yesterday's edit. The only thing that is missing is the $\sigma$-algebra, which is of course obvious in this context. There is no misery left. $\endgroup$ Jun 18, 2020 at 19:10

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