Let $(X,\tau)$ be a topological space and $\mathscr{B}(\tau)$ its Borel $\sigma$-algebra.
Question: Under which conditions does a measure $\mu:\mathscr{A}\rightarrow[0,\infty]$, defined on a sub-$\sigma$-algebra $\mathscr{A}\subset \mathscr{B}(\tau)$, extend to a Borel measure $\hat \mu:\mathscr{B}(\tau)\rightarrow [0,\infty]$? In particular I am interested in the case where $\mathscr{A}=\mathscr{B}(\tau')$ for some coarser topology $\tau'\subset \tau$.
If $\tau$ is generated by a metric $d$, then one approach is the following: Define $\mu^*:2^X\rightarrow [0,\infty]$ by $$ \mu^*(A)=\lim_{\delta \rightarrow 0} \left[ \inf\left\{\sum_{i=1}^\infty \mu(E_i): A\subset \bigcup_{i=1}^\infty E_i, E_i\in \tau, \mathrm{diam}_d(E_i)<\delta\right\} \right],$$ then according to a standard result in measure theory, $\mu^*$ is a metric outer measure and its restriction $\hat \mu = \mu^*\restriction\mathscr{B}(\tau)$ is a measure. In order for $\hat \mu$ to be an extension of $\mu$ it would thus suffice to show that $\mu=\mu^*\restriction \mathscr{A}$. Are there circumstances (in particular in the case that $\mathscr{A}=\mathscr{B}(\tau')$) in which this is possible to check easily?