Problem: Define the distance function between $(x_1, y_1)$ and $(x_2, y_2)$ for two points (where $x_i, y_i$ are real numbers) in the plane to be$$|y_1 - y_2| \text{ if } x_1=x_2; \quad 1 + |y_1- y_2| \text{ if } x_1 \ne x_2.$$
$d$ is indeed a metric, and $(𝑋,d)$ is a Locally compact Hausdorff space. Also, for any $f\in C_c(X)$, there's only finitely many $x_i$ such that there exists a $y$ with $f(x_i, y) \ne 0$, denoted $(x_1, \dots, x_m)$.
(These serve as previous subparts of this problem, which are resolved. For locally compactness, we can let $\overline{B_{1/2}(x, y)}$ to be a compact neighbourhood of $(x, y)$ since it only contains $\{(x, y'): |y' - y| \leq 1/2\}$, and compactness follows from any sequence in this set has a convergent subsequence. Also, finiteness of the points $x_i$ also follows from the compactness, since if not, then we may construct a sequence in the support of $f$ such that $d((x_i, y_i),(x_j, y_j)) \geq 1$ for all $i\ne j$.)
Here comes my question.
For $f$ above define $I(f) = \sum_{i=1}^{n}\int_{-\infty}^{\infty} f(x_i, y) dy$. Then $I(f)$ induces a Random measure $\mu$ on $X$. Is $\mu$ inner regular for all Borel set?
Here's my effort: Clearly, $\nu(\overline{B_{\delta}(x, y)}) = 2\delta$, for $\delta < 1/2$. Therefore, for the set $\{(x, y'): y\in E\subset \mathbb{R}\}$, we have inner regularity on this set since lesbegue measure is inner regular. For a general set, if the measure of $E$ is finite, then there'are at most countably many $x$ such that $E^{x} := \{y: (x, y')\in E\}$ has inner regulaity, and the rest follows from the previous case. For case where the measure is infinite, we should also have inner regularity. Therefore, I think we have the result that it is inner regular.