Let $p: Y \to X$ be a continuous open surjection between 2nd countable, locally compact, Hausdorff spaces. Suppose that for each $x \in X$ there is a positive Radon measure $\lambda^x$ on $Y$ such that:
$supp(\lambda^x) = p^{-1}(x)$; and
for any $f \in C_c(Y)$ the function $x \mapsto \int_{Y} f \, d\lambda^x$ belongs to $C_c(X)$.
Is it true that for any compact $K \subseteq Y$ we have $\sup_{x \in X} \lambda^x(K) < \infty$?
Potentially useful facts:
Since each $\lambda^x$ is Radon we have $\lambda^x(K) < \infty$ for all $x \in X$.
The function $x \mapsto \int_{Y} \chi_{K} \, d\lambda^x$is not nessesarily continous since $\chi_K$ is not nessesarily continous (here $\chi_K$ denotes the indicator function on $K$.)
I have a suspicion it may only be true for compact sets which are the closure of open sets.
This actually ended up being pretty easy.
It follows from Theorem 2.2 of Rudin's Real and Complex Analysis that there exists $f \in C_c(Y)$ such that $f|_{K} = 1$ and $f \ge \chi_K$. Since $f \in C_c(Y)$ we have that $x \mapsto \int_{Y}f \,d\lambda^x $ belongs to $C_c(X)$, and so $\sup_{x \in X} \int_{Y}f \,d\lambda^x < \infty$. It now follows that $$ \lambda_x(K) = \int_{Y} \chi_K \,d\lambda^x \le \int_{Y} f \,d\lambda^x < \infty. $$