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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?

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This is not true as stated (when I get to the office I'll look up the problem in Folland... Yes, that's what he says. Wouldn't be the first time he was wrong. Wouldn't be the first time I was wrong either...) There's a correct example of the phenomenon in question at the the bottom.

Say $X$ is the original space and $Y$ is $X$ minus the $y$-axis. Any compact subset of $Y$ is also a compact subset of $X$, so the restriction is finite on compact sets. Any open subset of $Y$ is open in $X$, so if the original measure is inner regular on open sets then so is the restriction.

If $E$ is a Borel subset of $X$ let $\mu_X^*(E)$ be the inf of $\mu(O)$ for $O$ an open set in $X$ containing $E$; similarly for $\mu_Y^*(E)$.

Suppose that $O$ is an open subset of $X$ and let $V=O\cap Y$. We certainly have $m_1(V^y)=m_1(O^y)$ for every $y$, and since $O$ is open it follows that $V^y=\emptyset$ if and only if $O^y=\emptyset$; hence $$\mu(V)=\mu(O).$$

This shows that if $E$ is a Borel subset of $Y$ then, assuming again that $\mu$ is actually Radon, $$\mu_Y^*(E)=\mu_X^*(E)=\mu(E).$$

Aha: Instead let $Z=\{0\}\times \Bbb R$. Now any subset of $Z$ is open, and if $O$ is an uncountable subset of $Z$ then $\mu(O)=\infty$. If $K\subset Z$ is compact then $K$ is finite; hence $\mu(K)=\sum_{y}m_1(K^y)=0$. So the restriction of $\mu$ to $Z$ is not inner regular on open sets.

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  • $\begingroup$ Thanks. I understand why your example at the bottom works, and I'll read over the rest of your answer and accept it soon. I re-read the exercise from Folland, and I believe that I've stated it correctly. Maybe it is a mistake. $\endgroup$ – user3281410 Feb 17 '18 at 18:53

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