Difference between revisions of "User:Boris Tsirelson/sandbox1"
From Encyclopedia of Mathematics
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| − | Every narrow neighborhood of a probability measure $\mu$ is also a wide neighborhood of $\mu$. Here is a proof. Given $\varepsilon$, we take a compactly supported continuous $f:X\to[0,1]$ such that $\int f \rd\mu > 1-\varepsilon$. Now, consider another probability measure $\nu$ widely close to $\mu$, namely, satisfying $|\int f \rd(\mu-\nu)| < \varepsilon$. If such $\nu$ satisfies $|\int fg \rd(\nu-\mu)|<\varepsilon$ | + | Every narrow neighborhood of a probability measure $\mu$ is also a wide neighborhood of $\mu$. Here is a proof. Given $\varepsilon$, we take a compactly supported continuous $f:X\to[0,1]$ such that $\int f \rd\mu > 1-\varepsilon$. Now, consider another probability measure $\nu$ widely close to $\mu$, namely, satisfying $|\int f \rd(\mu-\nu)| < \varepsilon$ and therefore $\int f \rd\nu > 1-2\varepsilon$. If such $\nu$ satisfies $|\int fg \rd(\nu-\mu)|<\varepsilon$ for a given $g:X\to[-1,1]$ then $|\int g \rd(\nu-\mu)|<C\varepsilon$. |
Revision as of 08:37, 9 August 2012
Every narrow neighborhood of a probability measure $\mu$ is also a wide neighborhood of $\mu$. Here is a proof. Given $\varepsilon$, we take a compactly supported continuous $f:X\to[0,1]$ such that $\int f \rd\mu > 1-\varepsilon$. Now, consider another probability measure $\nu$ widely close to $\mu$, namely, satisfying $|\int f \rd(\mu-\nu)| < \varepsilon$ and therefore $\int f \rd\nu > 1-2\varepsilon$. If such $\nu$ satisfies $|\int fg \rd(\nu-\mu)|<\varepsilon$ for a given $g:X\to[-1,1]$ then $|\int g \rd(\nu-\mu)|<C\varepsilon$.
How to Cite This Entry:
Boris Tsirelson/sandbox1. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Boris_Tsirelson/sandbox1&oldid=27458
Boris Tsirelson/sandbox1. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Boris_Tsirelson/sandbox1&oldid=27458