# Boundedness of the measure of a compact set under a continuous family of measures

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.

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