Is the support of a Borel measure measured the same as the whole space? Wikipedia says

Let (X, T) be a topological space. Let μ be a measure on the Borel σ-algebra on X. Then the support
  (or spectrum) of μ is defined to be the set of all points x in X for
  which every open neighbourhood Nx of x has positive measure.

The support of μ is a Borel-measurable subset, because

The support of a measure is closed in X.

I wonder if the support of μ is measured the same as the whole space?
It is equivalent to that the complement of the support of μ has 0 measure. But the following property seems to say it is not true

Under certain conditions on X and µ, for instance X being a
  topological Hausdorff space and µ being a Radon measure, a measurable
  set A outside the support has measure zero

So when does the support of a measure on a Borel sigma algebra have different measure from the whole space?
Thanks!

For an example with a probability measure, consider the following standard counterexample: let $X = [0, \omega_1]$ be the uncountable ordinal space (with endpoint), with its order topology.  This is a compact Hausdorff space which is not metrizable.  Define a probability measure on the Borel sets of $X$ by taking $\mu(B) = 1$ if $B$ contains a closed uncountable set, $\mu(B)=0$ otherwise.  It is known that this defines a countably additive probability measure; see Example 7.1.3 of Bogachev's Measure Theory for the details.
If $x \in X$ and $x < \omega_1$, then $[0, x+1)$ is an open neighborhood of $x$ which is countable, hence has measure zero.  So $x$ is not in the support of $\mu$.  In fact the support of $\mu$ is simply $\{\omega_1\}$.  But $\mu(\{\omega_1\}) = 0$.
