# strong convergence sufficient conditions

Would it be true that $\mu_n \to \mu$ strongly if $\int f\mathrm{d}\mu_{n}\to \int f\mathrm{d}\mu$ for every uniformly continuous function? Assume the space is $\mathbb{R}^{N}$ and has the usual topology. I am reading some paper and it seems to use this as a fact. I cannot see the difference between assuming $\mu_n \Rightarrow \mu$ and this condition mentioned above (uniformly continuous function). $\Rightarrow$ (weak convergence) obviously does not imply strong convergence, and so I am confused. The paper which I am talking about is "Weak convergence of a sequence of Markov Chains" - AF Karr, it is not a HW problem.

• By strong convergence I guess you mean $\mu_n(A)\to \mu(A)$ for each measurable $A$. If we take $\mu_n:=\delta_{n^{-1}}$ and $\mu:=\delta_0$, then $\int f\mathrm d\mu_n\to \int f\mathrm d\mu$ for each continuous $f$. But we don't have the strong convergence. – Davide Giraudo Sep 21 '13 at 8:31
• Do you have a link to the paper? I didn't manage to find it. – Davide Giraudo Sep 21 '13 at 15:34
• here it is link.springer.com/article/10.1007%2FBF00539859 its on page 44 (of the journal or same as page 4 of paper ) extreme bottom (6) Theorem ...... i can send you the paper in mail or message...it is "Weak Convergence of a Sequence of Markov Chains" Alan F. Karr – user24367 Sep 21 '13 at 19:53
• I would be thankful (my mail address in on my web page). – Davide Giraudo Sep 21 '13 at 20:02
• I got the point now, what I was saying was wrong.He used the fact that continuous functions with compact support (bounded continuous functions) are dense in $L^{1}$ to show that $\mu = \mu K$ via sequence of measures which converges weakly to $\mu$. $\mu(g) - \mu K(g) = \mu(g)-\mu^{n}(g)+\mu^{n}(g)-\mu K(g) =\mu(g)-\mu^{n}(g)+\mu^{n}K^{n}(g)-\mu K(g)$, where $\mu^{n}K^{n} = \mu^{n}$ continuing $\mu(g)-\mu^{n}(g)+\mu^{n}K^{n}(g)-\mu K(g)= \mu(g)-\mu^{n}(g)+\mu^{n}K^{n}(g)-\mu^{n}K(g)+\mu^{n}K(g)-\mu K(g)$, now if $K^{n}\to K$ pointwise and $\mu^{n}\Rightarrow \mu$ then the above goes to 0. – user24367 Sep 21 '13 at 21:09

The fact that $\int f\mathrm d\mu_n\to\int f\mathrm d\mu$ for each uniformly continuous function does not imply strong convergence, as $\mu_n:=\delta_{\frac 1n}$ and $\mu=\delta_0$ shows. This only gives weak convergence. And it is not what the author uses, as the OP noticed.
Strong convergence is equivalent to $\lim_n\int f\mathrm d\mu_n=\int f\mathrm d\mu$ for each $f\in L^\infty$, which is restrictive. Weak convergence is weaker than convergence in probability.