For, if the restriction of $\mu$ to an open set $U$ is zero, then so is that of $|\mu|$ (resp. of $\mu^+$ and $\mu^-$ when $\mu$ is real), and conversely.

Note that the supports of $\mu^+$ and $\mu^-$ can be nonempty and identical (cf. Ch. V, § 5, Exer. 5).

For, if the restriction of $\nu$ to an open set is zero, then so is that of $\mu$.

For, if the restrictions of two measures to an open set are zero, then the same is true of the restriction of their sum.

If $\mu$ and $\nu$ are two positive measures, then the support of $\lambda = \mu + \nu$ is equal to the union of the supports of $\mu$ and $\nu$; for, if $x_0$ is a point of the union and $V$ is any neighborhood of $x_0$, then there exists a continuous function $f \geq 0$ with support contained in $V$ and such that one of the numbers $\mu(f)$, $\nu(f)$ is $> 0$; a fortiori, $\lambda(f) = \mu(f) + \nu(f) > 0$.

The proposition is obvious from the definitions.

For, it is the intersection of the vaguely closed hyperplanes with equation $\mu(f) = 0$, where $f$ runs over the set of functions in $\mathscr{H}(X; \mathbf{C})$ whose support does not intersect $F$.

Suppose $X$ is not compact: given a filter $\Phi$ on the space $\mathscr{M}(X; \mathbf{C})$ of measures on $X$, we shall say that the support of a measure $\mu$ recedes indefinitely along $\Phi$ if, for every compact subset $K$ of $X$, there exists a set $M \in \Phi$ such that $\mathrm{Supp}(\mu) \cap K = \varnothing$ for every measure $\mu \in M$.

Let $f$ be any function in $\mathscr{K}(X; \mathbf{C})$ and let $K$ be its support. By hypothesis, there exists a set $M \in \Phi$ such that for every measure $\mu \in M$, $\operatorname{Supp}(\mu) \cap K = \varnothing$; it follows that $\mu(f) = 0$ for all $\mu \in M$, which proves the proposition.

Set $K = \operatorname{Supp}(f)$, $S = \operatorname{Supp}(\mu)$. Given a number $\varepsilon > 0$, let $V$ be the set of $x \in X$ such that $|f(x)| < \varepsilon$; $V$ is an open set containing $S$ by hypothesis, therefore $CS$ is a neighborhood of the compact set $CV$. It follows that there exists a continuous mapping $h$ of $X$ into $[0, 1]$, equal to 1 on $CV$ and with support contained in $CS$ (§ 1, No. 2, Lemma 1). Since the support of $fh$ does not intersect $S$, $\mu(fh) = 0$. On the other hand, $f = fh$ on $K \cap CV$, and $|fh| \leq |f|$ on $X$, therefore $|f - fh| \leq 2\varepsilon$ on $X$, by the choice of $V$. Observe finally that there exists a number $M_K$ such that $|\mu(g)| \leq M_K \|g\|$ for every function $g \in \mathscr{K}(X; \mathbf{C})$ whose support is contained in $K$; since the support of $f - fh$ is contained in $K$, $|\mu(f - fh)| \leq 2M_K \varepsilon$ and consequently $|\mu(f)| = |\mu(f - fh)| \leq 2M_K \varepsilon$; since $\varepsilon$ is arbitrary, $\mu(f) = 0$.

For, $f = |f|$ on $\operatorname{Supp}(\mu)$, therefore $\mu(f) = \mu(|f|) \geq 0$ by Corollary 1.

For, $\operatorname{Supp}(|\mu|) = \operatorname{Supp}(\mu)$, and if $h$ is a continuous mapping of $X$ into $[0, 1]$, equal to 1 on $\operatorname{Supp}(f)$ and with compact support, then $|f(x)| \leq ah(x)$ on $\operatorname{Supp}(\mu)$, therefore

by Corollary 2; the conclusion then follows from formula (13) of § 1, No. 6.

Let $x$ be a point of $X$ such that $f(x) > 0$; let us show that $x$ does not belong to $\mathrm{Supp}(\mu)$. For, there exist a compact neighborhood $V$ of $x$ and a number $a > 0$ such that $f(y) \geq a$ on $V$. If $g$ is any continuous function $\geq 0$ with support contained in $V$, let us show that $\mu(g) = 0$; indeed, if one sets $b = \|g\|$ then $g \leq bf/a$, whence $\mu(g) \leq b\mu(f)/a = 0$.

Set $S = \mathrm{Supp}(\mu)$; let $x_0$ be a point not belonging to $T$; there exists an open neighborhood $V$ of $x_0$ such that at every point of $V \cap S$, $g$ is zero; if $f \in \mathscr{K}(X; \mathbf{C})$ has support contained in $V$, then $fg$ is zero on $S$, therefore (Prop. 8) $\mu(gf) = 0$; in other words, the restriction of $g \cdot \mu$ to $V$ is zero.

Conversely, assuming that the restriction of $g \cdot \mu$ to an open neighborhood $W$ of a point $x_0 \in X$ is zero, let us show that there does not exist a point of $W \cap S$ at which $g$ is $\neq 0$. Indeed, if there were such a point $y$, there would exist a compact neighborhood $U$ of $y$, contained in $W$, at every point $x$ of which $g(x) \neq 0$; but then every function $f \in \mathscr{K}(X; \mathbf{C})$ with support contained in $U$ could be written $f = gh$, where $h \in \mathscr{K}(X; \mathbf{C})$ has support contained in $U \subset W$; it would then follow that $\mu(f) = \mu(gh) = 0$, contrary to the hypothesis $y \in S$.

Note that $T$ is contained in the intersection of the support $S$ of $\mu$ and the support of $g$, but it is not necessarily equal to this intersection. For example, if $X = \mathbf{R}$, $\mu$ is the Dirac measure at the point 0, and $g(x) = x$, then $g \cdot \mu = 0$ even though the intersection of the supports of $g$ and $\mu$ reduces to the point 0, thus is nonempty.

For, $|\mu|$ is also a measure with compact support, thus we can restrict attention to the case that $\mu \geq 0$; if $h$ is a continuous mapping of $X$ into $[0, 1]$, with compact support and equal to 1 on $\mathrm{Supp}(\mu)$, then for every function $f \in \mathscr{K}(X; \mathbf{C})$ one has $|f(x)| \leq \|f\| h(x)$ on $\mathrm{Supp}(\mu)$, therefore (Cor. 2 of Prop. 8) $\mu(|f|) \leq \mu(h)\|f\|$, which proves the proposition (§ 1, No. 8).

For, such a measure $\mu$ is zero for every function $f \in \mathscr{K}(X; \mathbf{C})$ satisfying the $n$ relations $f(a_i) = 0$ (No. 3, Prop. 8); since these relations may be written $\varepsilon_{a_i}(f) = 0$, $\mu$ is a linear combination of the $\varepsilon_{a_i}$ (A, II, §7, No. 5, Cor. 1 of Th. 7).

In particular, every measure whose support is either empty or reduced to a single point $x$ is of the form $\alpha \varepsilon_x$, where $\alpha$ is a complex number; such a measure is said to be a point measure; thus every measure whose support is finite is a sum of point measures.

It suffices to prove that $\mu$ is orthogonal to the subspace $V^\circ$ of $\mathscr{K}(X; \mathbf{C})$ orthogonal to $V$ (TVS, II, §6, No. 3, Cor. 2 of Th. 1), that is, that the relations $\langle f, \varepsilon_a \rangle = 0$, where $a$ runs over the support of $\mu$, imply $\langle f, \mu \rangle = 0$; but this is just Prop. 8 of No. 3.

To prove the first assertion, it suffices to establish that the measure $\mu$ belongs to the polar set of the polar set $A^\circ$ of $A$ in $\mathscr{K}(X; \mathbf{C})$ (TVS, II, §6, No. 3, Th. 1 and §8, No. 4); this means that for $f \in \mathscr{K}(X; \mathbf{C})$, the relations $|\langle f, \varepsilon_a \rangle| \leq 1/\| \mu \|$ for all $a \in \mathrm{Supp}(\mu)$ imply that $|\langle f, \mu \rangle| \leq 1$; but this is a consequence of Cor. 3 of Prop. 8 of No. 3.

To prove the second assertion, we note that

since the function $\nu \mapsto \| \nu \|$ is lower semi-continuous for the vague topology (§1, No. 9, Cor. 4 of Prop. 15), and the conclusion follows from the fact that $\| \nu \| \leq \| \mu \|$ for $\nu \in A$ by definition.

We can suppose that $\mu \neq 0$. Let $V$ be an open neighborhood of 0 for the vague topology; for every $\varepsilon$ such that $0 < \varepsilon < 1$, there exists, by virtue of Cor. 1, a measure $\nu_0$ whose support is finite and contained in $\mathrm{Supp}(\mu)$ and for which $\nu_0 - \mu \in V$ and $\| \mu \| \geq \| \nu_0 \| \geq (1 - \varepsilon) \| \mu \|$. Setting $\nu = (\| \mu \| / \| \nu_0 \|) \nu_0$, we have $\| \nu \| = \| \mu \|$ and $\| \nu - \nu_0 \| \leq \| \mu \|$; for $\varepsilon$ sufficiently small we therefore have $\nu - \mu \in V + V$, whence the conclusion.

The same reasoning as in Cor. 2 shows that we can limit ourselves to proving that $\mu$ is in the vague closure of the convex set $B$ formed by the positive measures with finite support contained in $\mathrm{Supp}(\mu)$ and with norm $\leq \| \mu \|$. Again, it suffices to establish that $\mu$ belongs to the polar set of $B^\circ$, the polar set of $B$ in the space $\mathscr{K}(X; \mathbf{R})$ (TVS, II, §6, No. 3, Th. 1); but this means that for $f \in \mathscr{K}(X; \mathbf{R})$ the relations $\langle f, \varepsilon_a \rangle \geq -1 / \| \mu \|$ for all $a \in \mathrm{Supp}(\mu)$ imply $\langle f, \mu \rangle \geq -1$, which is a consequence of No. 3, Cor. 2 of Prop. 8.

On the cone $\mathscr{M}_+(X)$, the topology of strictly compact convergence is identical to the vague topology (§1, No. 10, Prop. 18), and every measure on $X$ may be written $\mu_1 - \mu_2 + i \mu_3 - i \mu_4$, where the $\mu_j$ ($1 \leq j \leq 4$) are positive measures; the conclusion therefore follows from Th. 1.

The condition is obviously sufficient by virtue of Prop. 6 of No. 2. To see that it is necessary, suppose $x_0 \in \mathrm{Supp}(\mu)$; consider a finite number of functions $f_k$ ($1 \leq k \leq n$) in $\mathscr{K}(X; \mathbf{C})$, and an arbitrary number $\delta > 0$; we are to prove that there exists a function $g \in \mathscr{K}(X; \mathbf{C})$ such that $\| g \cdot \mu \| \leq 1$ and such that

for $1 \leq k \leq n$. Let $U$ be a relatively compact open neighborhood of $x_0$ such that the oscillation of each of the $f_k$ ($1 \leq k \leq n$) on $U$ is $\leq \delta / 2$. By hypothesis, since $x_0 \in \mathrm{Supp}(\mu)$, there exists a function $g_0 \in \mathscr{K}(X; \mathbf{C})$ whose support is contained in $U$ and such that $\mu(g_0) \neq 0$; the measure $\nu = g_0 \cdot \mu$ is not zero, since for every function $f \in \mathscr{K}(X; \mathbf{C})$ equal to 1 on $U$, $\nu(f) = \mu(g_0) \neq 0$. Moreover, $\nu$ is bounded (No. 3, Prop. 11);

multiplying $g_0$ by a scalar, we can suppose that $\| \nu \| = 1$. This being so, setting $\alpha_k = f_k(x_0)$ we can write, for $1 \leq k \leq n$ and for every function $h \in \mathscr{K}(X; \mathbf{C})$,

Since $\nu$ has its support in $U$, we may identify it with its restriction to $U$; the hypothesis $\| \nu \| = 1$ then implies that there exists a function $h \in \mathscr{K}(X; \mathbf{C})$, with support contained in $U$, such that $\| h \| \leq 1$ and such that $| \alpha_k (1 - \nu(h)) | \leq \delta / 2$ for $1 \leq k \leq n$. The definition of $U$ moreover shows that $| (\alpha_k - f_k(x)) h(x) | \leq \delta / 2$ for all $x \in U$; since $\| \nu \| = 1$ and $\operatorname{Supp}(\nu) \subset U$ we therefore have $| \nu((\alpha_k - f_k)h) | \leq \delta / 2$ and so, setting $g = g_0 h$,

This proves the proposition, since $\| g \cdot \mu \| = \| (g_0 h) \cdot \mu \| \leq \| g_0 \cdot \mu \| = 1$.

For, $\operatorname{Supp}(g \cdot \mu) \subset \operatorname{Supp}(\mu)$ by No. 3, Prop. 10; therefore the support of every vague limit of measures of the form $g \cdot \mu$ is also contained in $\operatorname{Supp}(\mu)$ (No. 2, Prop. 6). Conversely, if $\operatorname{Supp}(\nu) \subset \operatorname{Supp}(\mu)$ then $\nu$ is the vague limit of measures with finite support contained in $\operatorname{Supp}(\mu)$ (Th. 1), hence is in the vague closure of the set of measures $g \cdot \mu$ by Prop. 13.

Let $\mu$ be a discrete measure on $X$, defined by the masses $h(x) \neq 0$ placed at the points $x$ of a discrete closed subspace $N$ of $X$; let us show that $\operatorname{Supp}(\mu) = N$. For every $a \in N$ and every neighborhood $V$ of $a$, there exists a function $f \in \mathscr{K}(X; \mathbf{C})$ with support contained in $V$, equal to 1 at the point $a$ and to 0 at the other points of $N$, whence $\mu(f) = h(a) \neq 0$. On the other hand if $b \notin N$ then there exists a neighborhood $W$ of $b$ not intersecting $N$; for every function $g \in \mathscr{K}(X; \mathbf{C})$ with support contained in $W$, we therefore have $\mu(g) = 0$, which proves that $b \notin \operatorname{Supp}(\mu)$.

Conversely, let $\mu$ be a measure such that $\operatorname{Supp}(\mu)$ is a discrete closed subspace $N$ of $X$. For every $a \in N$, there exists an open neighborhood $V_a$ of $a$ that contains no point of $N$ distinct from $a$; the restriction of $\mu$ to $V_a$ is therefore a point measure with support $\{a\}$ (No. 2, Prop. 5), hence (No. 4, Prop. 12) is of the form $h(a)\varepsilon_a$, where $h(a) \neq 0$. Setting $h(x) = 0$ at the points of $\mathbf{C}\mathbf{N}$, and denoting by $\nu$ the measure defined by the masses $h(x)$, the principle of localization shows that $\nu = \mu$.

We thus see that on a discrete space $X$, every measure is discrete.