Measurable function with parameters

Question

Let $\mu_{y}$ a Borel probability on $\mathbb{R}^{n}$ indexed by a parameter $y \in \mathbb{R}^{d}$.

Let $A$ a borel set of $\mathbb{R}^{n}$ and suppose that the map $$ F \text{ : } y \rightarrow \mu_{y}(A) $$ is measurable (we equiped $\mathbb{R}$ with the borel algebra).

Let $T_{y} \text{ : } \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ a family of measurable functions indexed by a parameter $y \in \mathbb{R}^{d}$.

My question is : Is the map $$ y \rightarrow \mu_{y}( T_{y} \in A) $$ measurable ?

Thanks and regards.

EDIT : We can suppose that $(x,y) \rightarrow T_{y}(x)$ is measurable if it helps. Maybe without this assumption we can say anything.

Answer 1

No. Take $n=d=1$, $A = [0,1]$, and $\mu_y$ to be the Lebesgue measure for each $y$. Then, $y \mapsto \mu_y$ is measurable (it is just constant). Let $N$ be a non-measurable subset of $\mathbb{R}$. Then, for $y \in N$, define $T_y: \mathbb{R} \to \mathbb{R}$ by $T_y(x) = x$, and for $y \not \in N$, define $T_y: \mathbb{R} \to \mathbb{R}$ by $T_y(x) = \frac{x}{100}$. Then, each $T_y$ is measurable but $\{y : \mu(T_y^{-1}([0,1])) \in [0,2]\} = N$ is non-measurable.