First of all, one should mention that not every triple (X,B,μ) (i.e., what is often called a measure space)
satisfies the property that its C*-algebra of bounded functions is a von Neumann algebra (= W*-algebra) or that the map L^∞→(L_1)* is an isomorphism.
One has to impose additional conditions to ensure that this is true.
If we do not assume such a condition, Joseph Van Name's answer solves the problem.
However, if one is interested in a condition that would guarantee
that (X,B,I) (the σ-ideal I is often known as a measure class) comes from a triple (X,B,μ) whose algebra of bounded
functions is a von Neumann algebra (and pretty much all of measure theory happens in this restricted setting because otherwise almost every nontrivial theorem would fail), then
an answer was given in 1951 by Irving Segal in his paper
“Equivalences of measure spaces”.
Segal proved that several seemingly unrelated properties are equivalent for a triple (X,B,I) (i.e., a measure space equipped with a measure class).
Two of these properties have already been named: L^∞(X,B,I) is a von Neumann algebra and the canonical map L^∞(X,B,I)→L^1(X,B,I)* is an isomorphism (i.e., the Riesz representation theorem is satisfied).
Among other equivalent properties one can find the Radon-Nikodym theorem,
the Hahn decomposition theorem, and the fact that the Boolean algebra B/I is complete.
The full list and additional details can be found in the answer https://mathoverflow.net/a/20820/402.
Such triples (X,B,I) are known as localizable measurable spaces
and they can be organized into a category whose morphisms
(X,B,I)→(Y,C,J) are equivalence classes of measurable maps X→Y such that
the preimages of elements of J belong to I (this true in the complete case,
in the general case a slightly more complicated definition is necessary,
as described in the link).
This category turns out to be contravariantly equivalent to the category
of commutative von Neumann algebras (i.e., a version of Gelfand duality is true for von Neumann algebras), which further stresses the
importance of the localizability condition.
More information about this category and further links can be found in this answer:
https://mathoverflow.net/a/49542/402.
With respect to the norm topology, the space of Radon measures on a domain $\Omega$ is not seperable. Indeed, for any two distinct points $x,y$ in $\Omega$, the Dirac measures $\delta_x$ and $\delta_y$ (where $\delta_x(f)=f(x)$) satisfy $\|\delta_x-\delta_y\|=2$
since you can always find a compactly supported smooth function $f$ with $f(x)=-f(y)=1=\|f\|_\infty$. Any metric space that contains uncountably many disjoint open balls cannot be seperable. Of course there are many subspaces of Radon measures that are seperable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is seperable.
The space of Radon measures on a domain $\Omega$ is seperable in the weak$^*$ topology (This is probably the remark you allude to have read somewhere.) Indeed, consider the countable set $M_Q$ of measures of the form $\sum_{x \in S} a_x \delta_x$ where the coefficients $a_x$ are rational and $S$ runs over finite sets of points with rational coordinates. This $M_Q$ is countable and weak$^*$ dense. Also the embedding of $L^1$ as space of measures with absolutely continuous to Lebesgue measure is dense, and this gives another proof of weak$^*$ seperability.
Best Answer
If $I$ is uncountable, then in space $l^2(I)$ no countable set of functionals separates points. Consequently, for any set $A$ in the sigma-algebra generated by these functionals [the Baire sets for the weak topology, see reference below], if $0 \in A$, then an entire subspace is contained in $A$. So all elements of this sigma-algebra are unbounded. Thus this sigma-algebra is not all of the norm-Borel sets.
My papers on measurability in Banach space:
Indiana Univ. Math. J. 26 (1977) 663--677
Indiana Univ. Math. J. 28 (1979) 559--579
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For gaussian measures in Banach space, you really want the example of Fremlin and Talagrand, "A Gaussian measure on $l^{\infty}$". Ann. Probab. 8 (1980), no. 6, 1192--1193. This gaussian measure on $l^\infty$ with the cylindrical sigma-algebra has total mass 1, yet every ball of radius 1 has measure 0.