Lemma 3

— Let F be the set of functions $f \in \mathscr{C}(X; \mathbf{C})$ having the following property: there exist an $\alpha \in I$ and a function $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$ such that $f = f_\alpha \circ p_\alpha$. Then F is a dense linear subspace of $\mathscr{C}(X; \mathbf{C})$.

We note first of all that if $g = g_\beta \circ p_\beta$ and $h = h_\gamma \circ p_\gamma$, where $g_\beta \in \mathscr{C}(X_\beta; \mathbf{C})$ and $h_\gamma \in \mathscr{C}(X_\gamma; \mathbf{C})$, then there exists an $\alpha \in I$ such that $\alpha \geq \beta$ and $\alpha \geq \gamma$, and therefore $p_\beta = p_{\beta \alpha} \circ p_\alpha$, $p_\gamma = p_{\gamma \alpha} \circ p_\alpha$, which shows that

belongs to F; one sees similarly that $gh \in F$; F is thus a C-subalgebra of $\mathscr{C}(X; \mathbf{C})$, which contains the constants and is such that the relation $f \in F$ implies $\overline{f} \in F$. Finally, if $x \neq y$ are two points of X, there exists an $\alpha \in I$ such that $p_\alpha(x) \neq p_\alpha(y)$, therefore there is a function $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$ such that $f_\alpha(p_\alpha(x)) \neq f_\alpha(p_\alpha(y))$. The conclusion therefore follows from the Stone–Weierstrass theorem (GT, X, §4, No. 4, Prop. 7).

The lemma having been established, let $\mu, \mu'$ be two measures on X such that $(p_\alpha)_*(\mu) = (p_\alpha)_*(\mu')$ for all $\alpha \in I$; this means that, for every $\alpha \in I$ and every function $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$, we have

that is, $\langle f_\alpha \circ p_\alpha, \mu \rangle = \langle f_\alpha \circ p_\alpha, \mu' \rangle$; in other words, $\mu$ and $\mu'$ coincide on the subspace F of $\mathscr{C}(X; \mathbf{C})$, which is dense by Lemma 3, therefore $\mu = \mu'$, which proves (i).

(ii) The relation (15) applied to $p_\alpha$ shows that if $\mu$ is the inverse limit of the inverse system $(\mu_\alpha)$, necessarily

for all $\alpha \in I$.

(iii) Suppose the $p_{\alpha \beta}$ are surjective; one knows that the same is then true of the $p_\alpha$ (GT, I, §9, No. 6, Prop. 8). Consider an inverse system of measures $(\mu_\alpha)$ and let us first show that there exists a linear form $\lambda$ on F (in the notations of Lemma 3) such that, for every $\alpha \in I$ and every $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$, $\lambda(f_\alpha \circ p_\alpha) = \mu_\alpha(f_\alpha)$. To that end, let $\beta, \gamma$ be two indices in I, and $f_\beta \in \mathscr{C}(X_\beta; \mathbf{C})$, $f_\gamma \in \mathscr{C}(X_\gamma; \mathbf{C})$ two functions such that $f_\beta \circ p_\beta = f_\gamma \circ p_\gamma$; then there exists an index $\alpha \in I$ such that $\alpha \geq \beta$ and $\alpha \geq \gamma$, therefore

since $p_\alpha$ is surjective, this implies $f_\beta \circ p_{\beta \alpha} = f_\gamma \circ p_{\gamma \alpha}$, therefore

but by definition the last relation may also be written $\mu_\beta(f_\beta) = \mu_\gamma(f_\gamma)$, whence our assertion.

This being so, suppose that there exists a finite number $a > 0$ such that $\| \mu_\alpha \| \leq a$ for all $\alpha \in I$; then, for every function $f_\alpha \in \mathscr{C}(X_\alpha; \mathbf{C})$,

since $p_\alpha$ is surjective. This shows that the linear form $\lambda$ is continuous on $F$, and it follows from Lemma 3 that $\lambda$ may be extended to a measure $\mu$ on $X$ such that $(p_\alpha)_*(\mu) = \mu_\alpha$ for all $\alpha \in I$, which proves (iii).

(iv) To prove the existence of $\mu$ it suffices, by (iii), to verify that the family of norms $(\| \mu_\alpha \|)$ is bounded. But $\| \mu_\alpha \| = \mu_\alpha(1)$ and, when $\alpha \leq \beta$, the relation $\mu_\alpha = (p_{\alpha \beta})_*(\mu_\beta)$ implies that $\mu_\alpha(1) = \mu_\beta(1)$; since $I$ is directed, the total masses of all the measures $\mu_\alpha$ are therefore equal, whence our assertion. Moreover, the subspace $F$ obviously satisfies the property (P) of §1, No. 7, Prop. 9, therefore the inverse limit measure $\mu$ of $(\mu_\alpha)$ is positive. Finally, the relation $\mu_\alpha = (p_\alpha)_*(\mu)$ shows as above that $\mu(1) = \mu_\alpha(1)$.