Borel measure determined uniquely on a base Let $(X,\tau)$ be a topological space, $\mathcal{B}$ the $\sigma$-algebra generated by $\tau$ on $X$.
Let $\mu$ be a Borel measure on $X$. Does the restriction of $\mu$ to a base for the topology $\tau$, uniquely determine $\mu$?
For example, if $X$ is a metric space and $\mu$ is a finite Borel measure, the restriction of $\mu$ to $\tau$ uniquely determines $\mu$ on $\mathcal{B}$.
 A: If $\mu:\mathcal{B}\to \mathbb{C}$ is a finite Borel measure on a topological space $(X,\tau)$ such that $\mu|_\tau = 0$, then it holds that $\mu \equiv 0$. Dynkin's $\pi$-$\lambda$ argument is usually exploited in the standard proof of this fact.
Let us define
$$
\mathcal{L} = \{E\in \mathcal{B}\;|\;\mu(E) = 0\}.
$$ Then, we can show that (i)$X\in \mathcal{L}$, (ii) $E\in \mathcal{L}$ implies $E^c\in \mathcal{L}$, and (iii) if $\{E_n\}\subset \mathcal{L}$ is a disjoint family, then $\bigcup_{n\in \mathbb{N}} E_n \in \mathcal{L}$. This proves $\mathcal{L}$ is a $\lambda$-system (or Dynkin system.) Since $\tau \subset\mathcal{L}$ is a $\pi$-system, it follows from Dynkin's argument that
$$
\mathcal{B} = \sigma(\tau) \subset \mathcal{L},
$$ as desired.
However, it seems that the fact that $\mu=0$ on a basis $\mathcal{V}$ of $\tau$ does not imply $\mu = 0$. (In fact, I'm not very familiar with the set theory, so if there are some errors in my argument below, please point out through the comment.) We can see this by the example (inspired by the comment of Henno): 
Let $X$ be $[0,\Omega)$ where $\Omega$ is the first uncountable ordinal and $\mathcal{V}$ be the family of all sets of the form $[0,\omega)$ where $\omega<\Omega$. We declare set $E\subset [0,\Omega)$ is open if $E$ is of the form $E= [0,\omega)$ for some $\omega\leq \Omega$. We can see that $\mathcal{V}$ is a basis. Define $\mu(E) = 1$ if $\Omega\in\overline{E}$ and $\mu(E)=0$ otherwise (where the closure is taken with respect to the usual order topology.) To guarantee that $\mu$ is a measure, we state the following proposition: if $E\in \mathcal{B}$ is such that for all $\omega<\Omega$, $[\omega,\Omega) \not\subset E$, then $E\subset [0,\omega')$ for some $\omega'<\Omega$. To show this, let
$$
\mathcal{L} = \{E\in \mathcal{B} \;|\; [\omega,\Omega) \not\subset E,\;\forall \omega<\Omega\Rightarrow E\subset [0,\omega')\text{ for some }\omega'<\Omega\}. 
$$We first see that $[0,\Omega) \in \mathcal{L}$. Secondly, assume $E\in \mathcal{L}$. Then, if $[\omega,\Omega) \not\subset E,\;\forall \omega<\Omega$, then $E\subset [0,\omega')$ for some $\omega'$ and hence $[\omega',\Omega )\subset E^c$. This shows $E^c$ does not satisfy the premise and thus $E^c\in \mathcal{L}.$ If it is the case that $[\omega,\Omega) \subset E$ for some $\omega$, then $E^c\subset [0,\omega)$ satisfies the conclusion. Thus $E^c \in \mathcal{L}$. Next suppose $\{E_n\}\subset \mathcal{L}$ is a disjoint family. Assume that $[\omega,\Omega) \not\subset\bigcup_n E_n  $ for all $\omega<\Omega$. Then by the assumption, each $E_n \subset [0,\omega_n)$ for some $\omega_n<\Omega$. Then we have that $\bigcup_n E_n\subset [0,\sup_n \omega_n)$ where $\sup_n \omega_n$ is a countable ordinal and hence $\sup_n \omega_n<\Omega$. This shows $\bigcup_n E_n \in \mathcal{L}$ in this case. Assume conversely that $[\omega,\Omega) \subset\bigcup_n E_n $ for some $\omega<\Omega$. Then the premise is not satisfied and $\bigcup_n E_n\in \mathcal{L}$ in this case also. These arguments show that $\mathcal{L}$ is a $\lambda$-system containing $\tau$. Thus we get $\mathcal{B} \subset \mathcal{L}$ by Dynkin's theorem. 
So far, we've shown that if $E\in \mathcal{B}$, then it holds
$$
[\omega,\Omega) \not\subset E,\;\forall \omega<\Omega\Rightarrow E\subset [0,\omega')\quad \cdots(*)
$$for some $\omega'<\Omega$. We are finally ready to show that $\mu$ is actually a measure. Assume $\{E_n\}\subset \mathcal{B}$ is a disjoint family. In case $\Omega \in \overline{\bigcup_n E_n}$, then by the contraposition of $(*)$, there is $\omega<\Omega$ such that $[\omega,\Omega)\subset \bigcup_n E_n$. Suppose $\Omega \notin \overline{E_n}$ for all $n$. Then, by $(*)$, $E_n \subset [0,\omega_n)$ and we get $\bigcup_n E_n \subset [0,\sup_n \omega_n)$ leading to contradiction. Thus there is $m$ such that $\Omega \in \overline{E_n}$. By the contraposition of $(*)$, there is $\omega''<\Omega$ such that $[\omega'',\Omega) \subset E_m$. Since $\{E_n\}$ is disjoint, $E_n \subset [0,\omega'')$ for all $n\neq m$. This shows $\mu(E_m) = 1$ and $\mu(E_n) = 0$ for all $n\neq m$, proving
$$
\mu(\bigcup_n E_n) =1= \sum_n \mu(E_n).
$$ In case $\mu(\bigcup_n E_n)=0$, it must be that $0\leq \mu(E_n)\leq \mu(\bigcup_n E_n)=0$ and this establishes $\mu(\bigcup_n E_n) = \sum_n \mu(E_n)$. Thus, $\mu$ is countably additive and becomes a positive Borel measure. This provides an example of $\mu=0$ on $\mathcal{V}$, while $\mu\neq 0$.
A: Even for a general metric space this need not hold in general: suppose we have a real-valued measurable cardinal, then we have a probability measure $\mu$ on the power set of $\mathbb{R}$ that is $0$ on all singletons. Then considering the  base of singletons and having the discrete topology on the reals, we have that $\mu \equiv 0$ on the base but not on $\tau = \mathscr{P}(\mathbb{R})$. 
So I think we need to assume (in the metric case) that $X$ is a separable metric space, because then we have a countable base (inside any base) and the sigma-algebra is countably generated and all sorts of unicity arguments do apply.
I think on $X = \omega_1$ with the order topology and the Dieudonné measure $\mu$, the standard base also has the property that all base elements have measure $0$ while the measure itself is not identically $0$. So for general spaces it surely fails, even without assumptions like real-valued measurables...
