# Probability measure and Hausdorff distance [closed]

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 CountJul 13 at 0:22

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• Any two closed sets can be supports of proabability measures. So the first question does not make sense. – Kavi Rama Murthy Jul 12 at 12:16
• By support we mean the smallest closed set $A$ such that every open neighbourhood of any point of $A$ has positive measure. – user688435 Jul 12 at 12:30
• 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 $0$? – Kavi Rama Murthy Jul 12 at 12:35
• 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? – user688435 Jul 12 at 12:50

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].