My first question was answered within minutes by Sangchul Lee, but after a week of asking people I am stuck again, and let me describe.
I will try to write out the standard definitions carefully.
Let $T$ be a set, and let $A(T)$ be a sigma algebra on $T$. A measure is defined to be a function $\mu:A(T) \to {\mathbb R} $ such that
$\left\{\matrix{\mu(\{\})=0, \cr \mu(a)\ge 0 \ \ \forall a \in A, \cr \mu(\cup_\lambda a_\lambda) = \sum_\lambda \mu(a_\lambda) \hbox{ when } a_\lambda \hbox{ is a countable disjoint family}.}\right.$
Two of the axioms of Brownian motion say that $B_t -B_s$ is normal with variance $\vert s-t\vert$ and that intervals are independent. The third axiom says that Brpwmoam motion "has continuous paths."
The first two axioms determine a measure on cylinder sets such as $\{B: 4<B_3<7 \ \& \ 9 < B_{12} < 10\}$. Arbitrary unions of cylinder sets are open subsets of the topological space $T={\mathbb R}^{[0,\infty)}$ in the product topology, where $[0,\infty)$ is given the discrete topology; and consistency of the first two axioms implies a numerical probability measure on the open subsets of $T.$ This extends to a measure $\mu: A(T)\to {\mathbb R}.$ These last two facts are Kolmogorov and Caratheorody's extension theorems, and the measure $\mu$ is called the "law" of Brownian motion.
Now I am going to try to apply the axiom which says "Brownian motion has continuous paths." We can let $\Omega = C([0,\infty), {\mathbb R})\subset T$ and now people in the field seem to assume that we have a measure, which is a function $A(\Omega)\to {\mathbb R}$ satisfying the axioms of a measure.
I cannot manage to construct any measure on $\Omega.$ Initially I thought that the axiom means that the discontinuous paths are a Borel set of measure zero; but Sangchul Lee kindly explained that this is not the right way of understanding the axiom. The discontinuous functions are not one of the Borel sets on $T.$ So we have to construct a measure on $\Omega$ in some other way.
We do have the restriction function $r: A(T)\to A(\Omega)$ coming from the inclusion $\Omega \subset T$ and we do have the measure which is the law, coming from the first two axioms, which is a function $\mu: A(T) \to {\mathbb R}.$ But we cannot form the composite or restriction $\mu \circ r$ because the domain of $\mu$ is the domain of $r$, not the codomain.
If we are going to be able to compose $\mu$ with $r$ we need to follow the rule that the domain of $\mu$ has to equal the codomain of $r.$
How am I supposed to compose two functions which share a domain? If the complement of $\Omega$ is not Borel, and if I am not supposed to compose the inclusion function with the actual measure on $T$ , what does it mean to say that Brownian motion "has" continuous paths?
The word "has" has annoyed me for years about this, I can understand an axiom saying "groups have an identity element," but saying "Brownian motion has continuous paths" is more like defining Lie groups by just saying "Lie groups have a topology." For me, something is missing from the definition.