Weak convergence of measures on non-metrizable spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:44:35Z http://mathoverflow.net/feeds/question/67562 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67562/weak-convergence-of-measures-on-non-metrizable-spaces Weak convergence of measures on non-metrizable spaces Ricky Demer 2011-06-12T05:28:49Z 2011-06-12T12:22:13Z <p>(ZF + Countable Choice) <br><br><br><br><br> Let $\langle X,\mathcal{T} \hspace{.06 in} \rangle$ be a second-countable Hausdorff space. Let $\mu$ be a Borel measure on $X$. <br> Let $\langle I,\leq_I \rangle$ be a <a href="http://en.wikipedia.org/wiki/Directed_set" rel="nofollow">directed set</a>, and let <code>$\{\mu_i : i\in I\}$</code> be a collection of Borel measures on $X$. <br> Are the following are equivalent? <br><br><br> $1. \qquad$ For all open subsets $U$ of $X$, $\hspace{.12 in} \mu(U) \hspace{.08 in} \leq \hspace{.08 in} \displaystyle\liminf_i \hspace{.06 in} \mu_i(U)$</p> <p>$2. \qquad$ For all closed subsets $C$ of $X$, $\hspace{.12 in} \displaystyle\limsup_i \hspace{.06 in} \mu_i(C) \hspace{.08 in} \leq \hspace{.08 in} \mu(C)$ <br><br><br><br> Assume $\delta$ is a <a href="http://en.wikipedia.org/wiki/Proximity_space" rel="nofollow">proximity relation</a> on $X$ that induces $\mathcal{T} \hspace{.05 in}$. $\hspace{.04 in}$ Are the following equivalent? <br><br><br> For all Borel subsets $A$ and $B$ of $X$, if $\quad A \; \; \delta \; \; (X-B) \quad$ is false, then <br><br> $3. \qquad \mu(A) \hspace{.08 in} \leq \hspace{.08 in} \displaystyle\liminf_i \hspace{.06 in} \mu_i(B)$ <br><br> $4. \qquad \displaystyle\limsup_i \hspace{.06 in} \mu_i(A) \hspace{.08 in} \leq \hspace{.08 in} \mu(B)$ <br><br><br> Is (1 and 2) equivalent to (3 and 4) ?</p> http://mathoverflow.net/questions/67562/weak-convergence-of-measures-on-non-metrizable-spaces/67577#67577 Answer by Gerald Edgar for Weak convergence of measures on non-metrizable spaces Gerald Edgar 2011-06-12T12:22:13Z 2011-06-12T12:22:13Z <p>Allowing infinite measures will defeat equivalence of 1 and 2. Say Hausdorff measures of various dimensions $&lt;1$ on $\mathbb R$. All nonempty open sets have measure $\infty$. But closed sets can have interesting values. </p> <p>On the other hand, for finite measures, then of course 1 and 2 are equivalent, by taking complements.</p>