Support of a measure and set where it is concentrated
I have some problems understanding the difference between the notion of support of a measure and saying that a measure is concentrated on a certain set.
If I have a $\sigma$-algebra and $\mu$ a measure, is it equivalent to ask the support of $\mu $ to be contained in a measurable set $A$ and $\mu$ to be concentrated on $A$?
$\endgroup$ 22 Answers
$\begingroup$The support of a measure is the intersection of all closed sets that have full measure. This means that if a measure $\mu$ is concentrated on a set $A$, the inclusion $\operatorname{supp}(\mu) \subset \overline{A}$ holds [if $\mu$ is at least $\sigma$-finite].
This is pretty much all you can say about the set-theoretic relation between those two sets. Take for example the Lebesgue-measure. It's support is $\mathbb{R}$, but it is concentrated on every set of the form $\mathbb{R} \setminus \{x\}$. You can remove any element from $\mathbb{R}$ and the Lebesgue-measure is still supported on this "narrowed" set. But the support is uniquely determined as $\mathbb{R}$.
$\endgroup$ $\begingroup$Some supplementary information regarding Dominik's answer.
1) For most authors and in most cases of practical interest, the definition of support given by Dominik is appropriate. However, be aware that some authors do not distinguish between supports of a measure $\mu$ and sets where the measure $\mu$ concentrates ---the latter are also called carriers. See Billingsley (1995/1976, p. 23) and Davidson (1994).
2) The characterization of the support of a measure in terms of the smallest closed set with full measure is not the most general one because it does not always exist (e.g., Dudley 2002/1989, p. 238). For a Borel space $X$, the most general definition that I know is the following: The support of $\mu $ is the complement of the union of the $\mu$-null open sets, i.e., $\mathop{\mathrm{supp}} \mu:=\left(\bigcup\{O \subset X: O \text{ open and }\mu(O)=0\}\right)^c $.
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