Define a Radon measure to be inner regular on open sets, outer regular on Borel sets, and finite on compact sets, as in Folland's Real Analysis. A restriction of a $\sigma$-finite Radon measure to a Borel measurable subspace is again a Radon measure. However, this is not necessarily true for a non-$\sigma$-finite measure. I am trying to understand the counterexample given in problem 7.13 of Folland.
$X = \mathbb{R}\times \mathbb{R}$ where the first copy of $\mathbb{R}$ has the usual topology and the second copy has the discrete topology. $\mu$ is the unique Radon measure on $X$ induced by the functional on $C_{c}(X)$ defined by $f\mapsto \sum_{y}\int f(x,y)dx$ where $dx$ denotes integration with the Lebesgue measure. According to an answer in the post Reconciling several different definitions of Radon measures, $\mu$ is given by
(1) If $E$ has only countably many nonempty horizontal slices $E^{y}$, then $\mu(E)$ is the sum of the Lebesgue measure of each slice $E^{y}$
(2) If there are uncountably many nonempty horizontal slices $E^{y}$, then $\mu(E) = \infty$
The claim is that if we restrict $\mu$ to $\mathbb{R}\backslash \{0\}\times \mathbb{R}$, it is no longer Radon. Why is this true?