Measure of sections - continuous? Suppose $X$ is some space with a regular probability measure $\mu$, and let $A \subseteq \mathbb{R}^{n} \times X$ be some open set (in the product topology).
Is the function that measures the sections, $f(t)=\mu(A^t)$, continuous? ($A^t=\{x \in X: (t,x)\in A\}$)
To avoid problematic sets, I also assume that the sections $A^t$ are homeomorphic. Otherwise we can have something like $X=[0,1], A=(\mathbb{R}^{n} \setminus \{0\}) \times [0,1]$.
 A: Let $X = [0,1]$ be equipped with the Lebesgue measure and let $A$ be the set in $\mathbb{R}\times[0,1]$ which is the union of the rectangles $\mathbb{R}\times (\frac{1}{3}, \frac{2}{3})$ and $(0,1) \times (\frac{1}{4}, \frac{3}{4})$. This is a union of open sets and hence open. Moreover, every section is an open interval and therefore homeomorphic to any other section.
The function $f(t)$ is however not continuous since $f(0) = \frac{1}{3}$ while $f(t) = \frac{1}{2}$ for all $t \in (0,1)$.
A: Correct me if I am wrong but here are some steps that I think should lead you to the result: we can show that the property is true on A generators of opens, and that the property still holds when expanding to more general opens if you see what I mean
Let's assume that A is of the form $]a, b[ x U$ where U is an open of X that is part of the generating family of the opens of X.
The result is true in this case : f is constant.
If A is now a union of opens like this. Take $\epsilon > 0$. There is a finite subset of these opens such that $f'(t) > f(t) - \epsilon$. (f'(t) = same definition as f(t) but on the subset)
f'(t) is constant too on a small ball around t that is contained in each of those subsets. So $f'(u) = f'(t) > f(t) - \epsilon$ for u close enough to t and $f(x) >= f'(x)$ for any x so true for u.
The result is true when we take the complement of any open A.
So the result is true on every open A.
