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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:

  1. $supp(\lambda^x) = p^{-1}(x)$; and

  2. 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.

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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 $$ \lambda_x(K) = \int_{Y} \chi_K \,d\lambda^x \le \int_{Y} f \,d\lambda^x < \infty. $$

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