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There question is from Folland Real Analysis Chapter 7 Exercise 12. Let $X=\mathbb{R}\times\mathbb{R}_{d}$,where $\mathbb{R}_{d}$ denotes $\mathbb{R}$ with the discrete topology. If $f$ is a function on $X$, let $f^{y}=f(x,y)$; and if $E\subset X$, let $E^{y}=\{x:(x,y)\in E\}$.

(b) Define a positive functional on $C_{c}(X)$ by $I(f)=\sum_{y\in R}\int f(x,y)dx$ and let $\mu$ be the associated Radon Measure on $X$. Then $\mu(E)=\infty$ for any $E$ such that $E^{y}\neq\emptyset$ for uncountable any $y$.

I think $E$ is not compact here, but I do not know it is open or closed. Or this is not important. Also, I do not know how to use $I(f)=\sum_{y\in R}\int f(x,y)dx=\int fd\mu$. By the way, $\int f(x,y)dx$ means Lebesgue Measure?

  • First note that unaccountably many non-zero numbers will accumulate at some number different from zero. Adding numbers that accumulate to something other than zero, necessarily diverges. – user647486 Apr 15 '19 at 02:41
  • Sorry! I fail to understand you – user387147 Apr 15 '19 at 03:18

2 Answers2

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Let $E$ be a Borel set such that $E^y \ne \emptyset$ for $y\in A$, where $A$ is an uncountable subset of $\mathbb R$. By the outer regularity of $\mu$, it suffices to show that any open set containing $E$ has infinite measure.
Let $U$ be an open set containing $E$. There is $V \subseteq U$ such that $V^y = (a_y, b_y)$ with $a_y < b_y$ for $y \in A$.
Define $f : X \rightarrow \mathbb R$ so that $f^y \in C_c(\mathbb R, [0,1])$ and $\emptyset \ne \mathrm{supp}(f^y)\subseteq (a_y,b_y)$ for $y \in A$. Then $\int f(x,y)\, dx > 0$ for $y \in A$, which implies $\sum_{y\in \mathbb R} \int f(x,y)\, dx = \infty$.
Fix $M > 0$. By above, there is a finite set $F\subseteq \mathbb R$ such that $\sum_{y\in F} \int f(x,y)\, dx > M$.
Let $V_0 := V \cap (\mathbb R \times F)$ and $f_0 := \left.f\right|_{V_0}$.
By the Folland Ch 7 Exercise 12 (a), $f_0 \in C_c(X)$, and so \begin{equation} \mu(V_0) = \int \chi_{V_0}\, d\mu \geq \int f_0 \, d\mu = I(f_0) = \sum_{y\in \mathbb R} \int f_0(x,y)\, dx = \sum_{y\in F} \int f(x,y)\, dx > M \end{equation} Therefore, $\mu(U) > M$ for any $M>0$, which implies $\mu(U) = \infty$.

Luke
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Note that $\mu(E)=I(\chi_E)=\sum_{y\in \mathbb R}m(E^y)$ and the sum can converge only if $S=\{y:E^y\neq \emptyset \}$ is countable. That is, if $S$ is uncountable, then the series (of positive terms) diverges to $\infty.$

Matematleta
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