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Let $\mu$ be a Borel measure on $X$ that satisfies a zero-one law (i.e. $\mu$ taks takes only values $0,1$) and has $\mu(X) = 1$. Then $$F: C(X) \to \mathbb{R},\; f \mapsto \int_X f\; d \mu$$ defines a ring homomorphism.

I'm not sure, if there are not even examples for $(X, \mu)$ such that $F$ isn't an evalution (of course in this case $X$ can't be a Q-space). At least, there are such examples for Baire measures. [Edit: See "Added 2" for an affirmative answer]

Added: To give an example, let $\omega_1$ be the first uncountable ordinal and let $X := [0,\omega_1]$ be the set of all ordinals $0 \le \alpha\le\omega_1$ considered as topological space with the order topology. Then $X$ is a compact Hausdorff space. Futhermore, it can be shown that if $B$ is a Borel set, then $$\mu(B) := \begin{cases} 1, \text{ if } B \textstyle \text{ has an unbounded, closed subset of } X \setminus \lbrace \omega_1 \rbrace \newline 0, \textstyle\text{ otherwise}\end{cases}$$ defines a Borel measure on $X$ (cf. Halmos, Measure Theory, Exercise 52.10).

Since $\mu$ is zero on finite sets, it's obviously no dirac Dirac measure.

Added 2: The example also shows that the integral operator for a $\lbrace 0,1 \rbrace$-valued Borel measure is in general no evaluation:

Let $X_0 := [0,\omega_1) \subseteq X$. Then $\mu_0 := \mu|X_0$ is a $\lbrace 0,1 \rbrace$-valued Borel measure on $X_0$ with $\mu_0(X_0) = 1$ and
$$F_0: C(X_0) \to \mathbb{R},\; f \mapsto \int_{X_0} f\; d\mu_0$$ is a ring homomorphism that is no evaluation.

For, let $x \in X_0$ and set $f(\alpha) := 1$ if $\alpha \le x$, $f(\alpha) := 0$ if $\alpha > x$. $f$ is continuous and since $\mu([\alpha_0+1,\omega_1)) = 1$, $f = 0$ almost everythere. Hence $F_0(f) = 0$ but $e_x(f) = f(x) = 1$. Consequently there is no $x \in X_0$ such that $F_0 = e_x$.

Also note that if $X$ is compact, then, by the Riesz representation theorem, the only non-zero Radon measures with values in $\lbrace 0,1 \rbrace$ are the Dirac measures.

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Let $\mu$ be a Borel measure on $X$ that satisfies a zero-one law (i.e. $\mu$ taks only values $0,1$) and has $\mu(X) = 1$. Then $$F: C(X) \to \mathbb{R},\; f \mapsto \int_X f\; d \mu$$ defines a ring homomorphism.

I'm not sure, if there are not even examples for $(X, \mu)$ such that $F$ isn't an evalution (of course in this case $X$ can't be a Q-space). At least, there are such examples for Baire measures.

Added: To give an example, let $\omega_1$ be the first uncountable ordinal and let $X := [0,\omega_1]$ be the set of all ordinals $0 \le \alpha\le\omega_1$ considered as topological space with the order topology. Then $X$ is a compact Hausdorff space. Futhermore, it can be shown that if $B$ is a Borel set, then $$\mu(B) := \begin{cases} 1, \text{ if } B \textstyle \text{ has an unbounded, closed subset of } X \setminus \lbrace \omega_1 \rbrace \newline 0, \textstyle\text{ otherwise}\end{cases}$$ defines a Borel measure on $X$ (cf. Halmos, Measure Theory, Exercise 52.10).

Since $\mu$ is zero on finite sets, it's obviously no dirac measure.

Also note that if $X$ is compact, then, by the Riesz representation theorem, the only non-zero Radon measures with values in $\lbrace 0,1 \rbrace$ are the Dirac measures.

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Let $\mu$ be a Borel measure on $X$ that satisfies a zero-one law (i.e. $\mu$ taks only values $0,1$) and has $\mu(X) = 1$. Then $$F: C(X) \to \mathbb{R},\; f \mapsto \int_X f\; d \mu$$ defines a ring homomorphism.

I'm not sure, if there are not even examples for $(X, \mu)$ such that $F$ isn't an evalution (of course in this case $X$ can't be a Q-space). At least, there are such examples for Baire measures.

Added: To give an example, let $\omega_1$ be the first uncountable ordinal and let $X := [0,\omega_1]$ be the set of all ordinals $0 \le \alpha\le\omega_1$ considered as topological space with the order topology. Then $X$ is a compact Hausdorff space. Futhermore, it can be shown that if $B$ is a Borel set, then $$\mu(B) := \begin{cases} 1, \text{ if } B \textstyle \text{ has an unbounded subset of } X \setminus \lbrace \omega_1 \rbrace \newline 0, \textstyle\text{ otherwise}\end{cases}$$ defines a Borel measure on $X$ (cf. Halmos, Measure Theory, Exercise 52.10).

Since $\mu$ is zero on finite sets, it's obviously no dirac measure.

Also note that if $X$ is compact, then, by the Riesz representation theorem, the only non-zero Radon measures with values in $\lbrace 0,1 \rbrace$ are the Dirac measures.

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