# 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.
$${ mu: a < int_X f d mu , for arbitrary real $$a 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!