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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!

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This may be of interest:… – Byron Schmuland Mar 1 '12 at 0:49
@ByronSchmuland: Thanks! Right now I feel it a little beyond my understanding, but who knows if this will change soon. – Tim Mar 1 '12 at 0:52
A good place to look. But here, where we do not assume the space is metrizable, there are counterexamples without needing measurable cardinals. For example, Nate's. – GEdgar Mar 1 '12 at 1:31
up vote 4 down vote accepted

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 $\mu(\{\omega_1\}) = 0$.

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Alternate version: $[0,\omega_1)$. Same measure. Now it is a countably additive probability measure on a locally compact Hausdorff space, but the support is empty. The measure is sigma-smooth, but not tau-smooth. – GEdgar Mar 1 '12 at 1:33

Let $T$ be the discrete topology on $\mathbb{R}$ (so that every set is open) and let $m$ be a measure such that $m(A) = 0$ if $A$ is countable and $m(A) = \infty$ if $A$ is uncountable, for $A \subseteq \mathbb{R}$.

Claim: this is actually a measure. There are two properties to check. First, the empty set is countable, and thus has measure zero, as required. Second, suppose that $(A_n : n \in \mathbb{N})$ is a sequence of disjoint subsets of $\mathbb{R}$. If all the sets are countable, then so is their union, and so $$m(\bigcup_{n \in \mathbb{N}} A_n) = \sum_{n \in \mathbb{N}} m(A_n) = 0$$ as desired. On the other hand, if at least one of the sets $A_n$ is uncountable, then so is their union, and so $$m(\bigcup_{n \in \mathbb{N}} A_n) = \sum_{n \in \mathbb{N}} m(A_n) = \infty$$ (recall $\infty +\infty = \infty$). This proves the claim.

Now we look at the the support of $m$. Every point has a countable neighborhood; in fact $\{x\}$ is a neighborhood of a point $x$. These neighborhoods all have measure zero, so the support of $m$ is empty. On the other hand the measure is not identically zero.

This sort of thing cannot happen on the usual topology of $\mathbb{R}$ because it is Lindelöf. With the usual topology, the complement of the support, which naturally has a covering by open sets of measure zero, must have a covering by a sequence of open sets of measure zero, and thus the complement of the support itself has measure zero, as does all of its subsets.

The Wikipedia article suggests different sufficient conditions for the measure to ensure that the support has measure zero: that the space is Hausdorff and the measure is Radon (locally finite and inner regular). The pathological measure I constructed above is locally finite and the space is Hausdorff, but the measure is not inner regular.

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Also note that, because the topology is discrete, the space itself is a complete metric space. – Carl Mummert Mar 1 '12 at 1:12

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