Support (measure theory)

In mathematics, the support (sometimes topological support or spectrum) of a measure ''μ'' on a measurable topological space (''X'', Borel(''X'')) is a precise notion of where in the space ''X'' the measure "lives". It is defined to be the largest ( closed) subset of ''X'' for which every open neighbourhood of every point of the set has positive measure.

Motivation

A (non-negative) measure $\mu$ on a measurable space $(X, \Sigma)$ is really a function \mu : \Sigma \to , +\infty. Therefore, in terms of the usual definition of support, the support of $\mu$ is a subset of the σ-algebra $\Sigma$:

:$\operatorname (\mu) := \overline,$

where the overbar denotes set closure. However, this definition is somewhat unsatisfactory: we use the notion of closure, but we do not even have a topology on $\Sigma$. What we really want to know is where in the space $X$ the measure $\mu$ is non-zero. Consider two examples:
# Lebesgue measure $\lambda$ on the real line $\mathbb$. It seems clear that $\lambda$ "lives on" the whole of the real line.
# A Dirac measure $\delta_p$ at some point $p \in \mathbb$. Again, intuition suggests that the measure $\delta_p$ "lives at" the point $p$, and nowhere else.

In light of these two examples, we can reject the following candidate definitions in favour of the one in the next section:
# We could remove the points where $\mu$ is zero, and take the support to be the remainder $X\setminus \$. This might work for the Dirac measure $\delta_p$, but it would definitely not work for $\lambda$: since the Lebesgue measure of any singleton is zero, this definition would give $\lambda$ empty support.
# By comparison with the notion of strict positivity of measures, we could take the support to be the set of all points with a neighbourhood of positive measure:

:::$\$

:(or the closure of this). It is also too simplistic: by taking $N_x = X$ for all points $x \in X$, this would make the support of every measure except the zero measure the whole of $X$.

However, the idea of "local strict positivity" is not too far from a workable definition.

Definition

Let (''X'', ''T'') be a topological space; let B(''T'') denote the Borel σ-algebra on ''X'', i.e. the smallest sigma algebra on ''X'' that contains all open sets ''U'' ∈ ''T''. Let ''μ'' be a measure on (''X'', B(''T'')). Then the support (or spectrum) of ''μ'' is defined as the set of all points ''x'' in ''X'' for which every open neighbourhood ''N''_''x'' of ''x'' has positive measure:

:$\operatorname (\mu) := \.$

Some authors prefer to take the closure of the above set. However, this is not necessary: see "Properties" below.

An equivalent definition of support is as the largest ''C'' ∈ B(''T'') (with respect to inclusion) such that every open set which has non-empty intersection with ''C'' has positive measure, i.e. the largest ''C'' such that:

:$(\forall U \in T)(U \cap C \neq \varnothing \implies \mu (U \cap C) > 0).$

Properties

* $\operatorname (\mu_1 + \mu_2) = \operatorname (\mu_1) \cup \operatorname (\mu_2)$
* A measure ''μ'' on ''X'' is strictly positive if and only if it has support supp(''μ'') = ''X''. If ''μ'' is strictly positive and ''x'' ∈ ''X'' is arbitrary, then any open neighbourhood of ''x'', since it is an open set, has positive measure; hence, ''x'' ∈ supp(''μ''), so supp(''μ'') = ''X''. Conversely, if supp(''μ'') = ''X'', then every non-empty open set (being an open neighbourhood of some point in its interior, which is also a point of the support) has positive measure; hence, ''μ'' is strictly positive.
* The support of a measure is closed in ''X'', as its complement is the union of the open sets of measure 0.
* In general the support of a nonzero measure may be empty: see the examples below. However, if ''X'' is a Hausdorff topological space and ''μ'' is a Radon measure, a measurable set ''A'' outside the support has measure zero:
::$A \subseteq X \setminus \operatorname (\mu) \implies \mu (A) = 0.$
: The converse is true if ''A'' is open, but it is not true in general: it fails if there exists a point ''x'' ∈ supp(''μ'') such that ''μ''() = 0 (e.g. Lebesgue measure).
: Thus, one does not need to "integrate outside the support": for any measurable function ''f'' : ''X'' → R or C,
::$\int_X f(x) \, \mathrm \mu (x) = \int_ f(x) \, \mathrm \mu (x).$
* The concept of ''support'' of a measure and that of spectrum of a self-adjoint linear operator on a Hilbert space are closely related. Indeed, if $\mu$ is a regular Borel measure on the line $\mathbb$, then the multiplication operator $(Af)(x)=xf(x)$ is self-adjoint on its natural domain

::$D(A)=\$

:and its spectrum coincides with the essential range of the identity function $x \mapsto x$, which is precisely the support of $\mu$.Mathematical methods in Quantum Mechanics with applications to Schrödinger Operators

Examples

Lebesgue measure

In the case of Lebesgue measure ''λ'' on the real line R, consider an arbitrary point ''x'' ∈ R. Then any open neighbourhood ''N''_''x'' of ''x'' must contain some open interval (''x'' − ''ε'', ''x'' + ''ε'') for some ''ε'' > 0. This interval has Lebesgue measure 2''ε'' > 0, so ''λ''(''N''_''x'') ≥ 2''ε'' > 0. Since ''x'' ∈ R was arbitrary, supp(''λ'') = R.

Dirac measure

In the case of Dirac measure ''δ''_''p'', let ''x'' ∈ R and consider two cases:
# if ''x'' = ''p'', then every open neighbourhood ''N''_''x'' of ''x'' contains ''p'', so ''δ''_''p''(''N''_''x'') = 1 > 0;
# on the other hand, if ''x'' ≠ ''p'', then there exists a sufficiently small open ball ''B'' around ''x'' that does not contain ''p'', so ''δ''_''p''(''B'') = 0.
We conclude that supp(''δ''_''p'') is the closure of the singleton set , which is itself.

In fact, a measure ''μ'' on the real line is a Dirac measure ''δ''_''p'' for some point ''p'' if and only if the support of ''μ'' is the singleton set . Consequently, Dirac measure on the real line is the unique measure with zero variance rovided that the measure has variance at all

A uniform distribution

Consider the measure ''μ'' on the real line R defined by
:$\mu (A) := \lambda (A \cap (0, 1))$
i.e. a uniform measure on the open interval (0, 1). A similar argument to the Dirac measure example shows that supp(''μ'') =  , 1 Note that the boundary points 0 and 1 lie in the support: any open set containing 0 (or 1) contains an open interval about 0 (or 1), which must intersect (0, 1), and so must have positive ''μ''-measure.

A nontrivial measure whose support is empty

The space of all countable ordinals with the topology generated by "open intervals" is a locally compact Hausdorff space. The measure ("Dieudonné measure") that assigns measure 1 to Borel sets containing an unbounded closed subset and assigns 0 to other Borel sets is a Borel probability measure whose support is empty.

A nontrivial measure whose support has measure zero

On a compact Hausdorff space the support of a non-zero measure is always non-empty, but may have measure 0. An example of this is given by adding the first uncountable ordinal Ω to the previous example: the support of the measure is the single point Ω, which has measure 0.

Signed and complex measures

Suppose that ''μ'' : Σ →  minus;∞, +∞is a signed measure. Use the Hahn decomposition theorem to write

:$\mu = \mu^ - \mu^,$

where ''μ''^± are both non-negative measures. Then the support of ''μ'' is defined to be

:$\operatorname (\mu) := \operatorname (\mu^) \cup \operatorname (\mu^).$

Similarly, if ''μ'' : Σ → C is a complex measure, the support of ''μ'' is defined to be the union of the supports of its real and imaginary parts.

References

*
* (See chapter 2, section 2.)
* (See chapter 3, section 2)

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