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Let $X$ be a compact metric space with probability measures $\gamma$ and $\eta$ with respect to the Borel sigma algebra on $X$. Let $A = \text{support of } \gamma$ and $B = \text{support of } \eta$. Prove or disprove that the Hausdorff distance between $A$ and $B$ is zero. In general, what can we say about the Hausdorff distance between $C$ and $D$ where these sets are closed subsets of $X$ and satisfies $\gamma (C) = 1 = \gamma (D)$.

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closed as off-topic by RRL, cmk, Adrian Keister, José Carlos Santos, The Count Jul 13 at 0:22

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  • $\begingroup$ Any two closed sets can be supports of proabability measures. So the first question does not make sense. $\endgroup$ – Kavi Rama Murthy Jul 12 at 12:16
  • $\begingroup$ By support we mean the smallest closed set $A$ such that every open neighbourhood of any point of $A$ has positive measure. $\endgroup$ – user688435 Jul 12 at 12:30
  • $\begingroup$ Yes,but in the first question you are taking two probability measures $\gamma $ and $\eta$. That makes $A$ and $B$ totally arbitrary. Why do you think that Hausdorff distance between any two closed sets is $0$? $\endgroup$ – Kavi Rama Murthy Jul 12 at 12:35
  • $\begingroup$ I am trying to find conditions where we can say Hausdorff distance is zero (in first question).. So I started with the support but couldn't find any answer. Can we put some conditions on measures to answer first question positively? $\endgroup$ – user688435 Jul 12 at 12:50
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Second question: let $X=[0,1]$. Let $\gamma =\delta_{\frac 1 2}$, $C=[0,\frac 1 4] \cup \{\frac 1 2\}$ and $D=[\frac 3 4,1] \cup \{\frac 1 2\}$. The $\gamma (C)=\gamma (D)=1$ and the Hausdorff distance between $X=C$ and $D$ is $\frac 3 4$. [As mentioned in my comment above the first part is false too].

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