If I have a second countable topological space X, can i Always find a finite borel measure, such that every non-empty open set has positive measure?
without second countability, the discrete topology on $\mathbb R$ is a counter example.
If I have a second countable topological space X, can i Always find a finite borel measure, such that every non-empty open set has positive measure?
without second countability, the discrete topology on $\mathbb R$ is a counter example.
A simple solution: if $X$ is second countable, let $D=\{d_n : n =1,2,3,\ldots\}$ be a dense subset of $X$ and define $$\mu(A)= \sum_{n:d_n \in A}\frac{1}{2^n}$$ for all subsets of $X$.
Then clearly $\mu(X)=1$ and $\mu(O)>0$ for all $O$ non-empty and open.
If you want an atomless measure, we need at least that $X$ is crowded, and then we must maybe assume some more on $X$, e.g. to avoid cases like $\mathbb{Q}$ which is second countable and crowded but all of whose measures are atomic by countability of the space.
See Corollary 2.8 in this paper:
If $X$ is perfect, compact and metrizable, then there is a non-atomic regular Borel measure of full support on $X$.