# A very natural question in weak* topology [closed]

Can you provide me a counter example for this.

Suppose that I have a sequence of probability measures $(\mu_{r,t})_{r,t>0}$ on a compact space metric $X.$

there exists a probability measure $\mu_{r,0}$ such that $\mu_{r,t}\to^{*,t\to 0}\mu_{r,0}$

and

there exists a probability measure $\mu_{0,0}$ such that $\mu_{r,0}\to^{*,r\to 0}\mu_{0,0}.$

Is it true that there exist the weak* limit of $\mu_{r,t}$ as $r\to 0$ for every $t$ close to $0?$

## closed as off-topic by Michael Renardy, Bill Johnson, Stefan Kohl, Ryan Budney, Steven SamMay 28 '14 at 0:24

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• It seems that your desired conclusion is the same as your first assumption. Could you please check? Maybe you meant to reverse t and r? – Nate Eldredge May 27 '14 at 14:03
• No. $\mu_{r,t}=\delta_{t/r}$ for $r,t>0$ (and $X=[0,1]$). – Jochen Wengenroth May 27 '14 at 14:19
• When $r<t$ this isn't a measure on $X$. – Ian Morris May 27 '14 at 14:23
• What about an example on $X=[0,1]$? – user39115 May 27 '14 at 14:30
• @user39115: how about $\delta_{\sin(t/r)}$? – Ian Morris May 27 '14 at 14:35

Let $X=\{1,2\}$ and define $\mu_{r,t}:= \delta_1$ either if $r \geq t$, or if both $r<t$ and $\lfloor \frac{1}{r}\rfloor$ is even. Define $\mu_{r,t}:=\delta_2$ if both $r<t$ and $\lfloor \frac{1}{r}\rfloor$ is odd. Clearly $\lim_{t \to 0}=\delta_1$ for every $r>0$, and $\lim_{r \to 0} \lim_{t \to 0}\mu_{r,t}=\delta_1$ as hypothesised. It is equally clear that $\lim_{r \to 0} \mu_{r,t}$ does not exist for any $t>0$.
There is a heuristic underlying this answer: the set of all probability measures on $X$ in the weak-* topology contains a homeomorphic copy of $X$ as a closed subset, embedded by the map $x \mapsto \delta_x$. Thus if one can find a sequence $(x_{r,t})$ of elements of $X$ with analogous convergence properties then the measures $(\delta_{x_{r,t}})$ will be a counterexample.