Fractals: Why do metric and weak topologies match for measurements?

Leave $ (X, d) $ be a complete metric space,

$$ M ^ 1: = { mu: mu mbox {is a regular measure of Borel that has bounded support and} mu (X) = 1 }, $$

Y $$ BC (X): = {f: f: X rightarrow mathbb {R} mbox {is continuous and bounded in delimited subset} }. $$

In Hutchinson's article "Fractals and self-similarity", it is stated that the following topology coincides with $ M ^ 1 cap { mu: mu mbox {has compact support} } $:

  1. The topology generated taking as sub-base all the sets of the form.
    $ { mu: a < int_X f d mu <b } $, for arbitrary real $ a <b $ and arbitrary $ f in BC (X) $.

  2. The topology generated by the $ L $ metric in $ M ^ 1 $, which is given by $ L ( mu, nu): = sup { int_X f d mu – int_X f d nu: f in BC (X), mbox {Edge} f leq 1 } $, where $ mbox {Lip} f: = sup_ {x neq y} [|f(x)-f(y)|/d(x,y)]$.

My question is why this is true. My focus and uncertainty is that I'm not sure if a function with unlimited functions $ mbox {Lip} $ (e.g. $ sqrt {x} $ in $[0,1]$) It can be approximated by functions with limits. $ mbox {Lip} $ (for example, polynomials in $[0,1]$) in a complete general metric space.

Thank you!