Proposition 10

— Let $\mu$ be a measure on a locally compact space $X$; for every function $g \in \mathscr{C}(X; \mathbf{C})$, the support of the measure $g \cdot \mu$ is the closure $T$ of the set of points $x \in \mathrm{Supp}(\mu)$ such that $g(x) \neq 0$.

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.