User:Boris Tsirelson/sandbox1

Every narrow neighborhood of a probability measure $\mu$ is also a wide neighborhood of $\mu$. Here is a sketch of 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 $\int f \rd\nu > 1-2\varepsilon$.

Claim: If such $\nu$ satisfies $|\int fg \rd(\nu-\mu)|<\varepsilon$ for a given $g:X\to[-1,1]$ then $|\int g \rd(\nu-\mu)|<4\varepsilon$.

Proof of the claim: $ |\int g \rd(\nu-\mu) - \int fg \rd(\nu-\mu)| = |\int (1-f)g \rd(\nu-\mu)| \le |\int (1-f)g \rd\nu| + |\int (1-f)g \rd\mu| \le \int (1-f) \rd\nu + \int (1-f) \rd\mu < 2\varepsilon + \varepsilon $.