construction of non measurable function

Let $P$ denote the set of probability meausres. Consider a map $f$ defined by $f\colon P\to \mathbb R$. Now lets consider the conditional distribution $X\mid Y=y$ for two real valued random variables $X,Y$ and calculate $f(P_{X\mid Y=y})$. I am curious if $y\mapsto f(P_{X\mid Y=y})$ has to be measurable.

Some measurability assumptions are needed on the function $f$.

Some technical considerations:

For simplicity assume $\mathcal{P}(\mathbb{R})$ is the collection of Borel probability measures in the real line $\mathbb{R}$. First one equips $\mathcal{P}(\mathbb{R})$ with a $\sigma$-algebra in order to be able to talk about measurable functions on $\mathcal{P}(\mathbb{R})$.

For any bounded (real-valued) measurable function $h$ on $\mathbb{R}$. define $\phi_h:\mathcal{P}\rightarrow\mathbb{R}$ as $$\phi_h(\mu)=\int_{\mathbb{R}} h\,d\mu$$

Since $\mathbb{R}$ is a locally compact and complete separable space there are several ways to put a measurable structure in $\mathbb{R}$.

Consider the $\sigma$-algebra $\mathscr{M}(\mathcal{P})$ as the minimal $\sigma$-algebra for which the maps $\phi_K:\mu\mapsto \int_{\mathbb{R}}\mathbb{1}_K\,d\mu=\mu(K)$, $K$ compact, are measurable. This is the $\sigma$-algebra on $\mathcal{P}$ generated by $$\mathscr{C}=\{\phi_K^{-1}(a,\infty): K\,\text{compact},\, a\in \mathbb{R}\}$$ It turns out that $\mathscr{M}(\mathcal{P})$ is also generated by $$\mathscr{F}:=\{\phi^{-1}_h(U): h\in\mathcal{C}^+_{00}(\mathbb{R}),\, U\in\mathscr{B}(\mathbb{R})\}$$ where $\mathcal{C}^+_{00}(\mathbb{R})$ is the collection of all nonnegative real-valued functions on $\mathbb{R}$ that have compact support. All this can be seen in books on Stochastic geometry. Kallenberg's book on Probability (reference below) has a short introduction to this.

Lemma: Suppose $(X,\mathscr{A})$ is a measurable space and let $g:X\rightarrow \mathcal{P}$ be a map. $g$ is $\mathscr{A}-\mathscr{M}(\mathcal{P})$ measurable iff $\phi_h\circ g:X\rightarrow\mathbb{R}$ is $\mathscr{A}-\mathscr{B}(\mathbb{R})$ measurable for any $h\in\mathcal{C}^+_{00}(\mathbb{R})$.

Proof: Necessity is obvious.
For sufficiency, suppose $\phi_h\circ g$ is measurable for all $h\in\mathcal{C}^+_{00}(\mathbb{R})$, and for any $U\in\mathscr{B}(\mathbb{R})$, $(\phi_h\circ g)^{-1}(U)=g^{-1}(\phi^{-1}_h(U))\in\mathscr{A}$. Then, $g^{-1}(M)\in\mathscr{A}$ for all $M\in\mathscr{M}(P)$ for the condition is valid for any set in $\mathscr{F}$.

Solution to the OP: Consider now a probability space $(\Omega,\mathscr{F},\mathbb{P})$ and $X:(\Omega,\mathscr{B})\rightarrow(\mathbb{R},\mathscr{B}(\mathbb{R}))$. Part of the disintegration theorem shows that for any $\sigma$-algebra $\mathscr{A}\subset \mathscr{F}$, there is a regular version of $\mathbb{P}[X\in\cdot|\mathscr{A})$, that is, there is a stochastic kernel $\mu:\Omega\times\mathscr{B}(\mathbb{R})\rightarrow\mathbb{R}$ such that

1. for each $U\in\mathscr{B}(\mathbb{R})$, the map $\omega\mapsto\mu(\omega,U)$ is a $\mathscr{A}$ measurable map,
2. There is a $\mathbb{P}$-negligible set $N$ such that for all $\omega\in \Omega\setminus N$, $\mathbb{P}[X\in U|\mathscr{A}](\omega)=\mu(\omega,U)$.

Claim: The map $G:\Omega\mapsto\mathscr{M}(\mathcal{P}(\mathbb{R}))$ given by $$\omega\mapsto \mu(\omega,\cdot)$$ is $\mathscr{A}-\mathscr{M}(\mathcal{P}(\mathbb{R}))$ measurable. Indeed, by the the disintegration theorem, for any $h\in\mathcal{C}_{00}(\mathbb{R})$ $$\mathbb{E}[f(X)|\mathscr{A}](\omega)=\int_{\mathbb{R}}h(x)\,\mu(\omega,dx),\qquad \omega\in\Omega\setminus N$$ The map $\omega\mapsto\int_{\mathbb{R}}h(x)\,\mu(\omega,dx)$ is $\mathscr{A}$-measurable. Therefore $G$ is measurable.

Finally, if $f:\mathscr{M}(\mathcal{P}(\mathbb{R}))\rightarrow(\mathbb{R},\mathscr{B}(\mathbb{R}))$ is a measurable function on probability measures, then $f\circ G$ is $(\Omega,\mathscr{A})-\mathscr{B}(\mathbb{R})$ measurable being the composition of measurable functions. $\Box$.

Example 1: In the case of the OP, one takes $\mathscr{A}=\sigma(Y)$, the $\sigma$-algebra generated by $Y$. In such case, the regular version of $\mathbb{P}[X\in dx|\sigma(Y)]$ can be expressed as $$\mathbb{P}[X\in dx|\sigma(Y)](\omega)=\mu'(Y(\omega),\cdot)$$ for some stochastic kernel $\mu':T\times\mathscr{B}(\mathbb{R})\rightarrow\mathbb{R}$ from $T$ to $\mathbb{R}$. The argument we presented implies that $\omega\mapsto f(\mathbb{P}[X\in dx|\sigma(Y)](\omega))=f(\mu'(Y(\omega),dx)$ is measurable.

Example 2: If $(\Omega,\mathscr{G},\mathbb{P})=(\mathbb{R}\times T,\mathscr{B}(\mathbb{R})\otimes\mathscr{T},P)$, where $P$ is a probability measure on $\mathscr{B}(\mathbb{R})\otimes\mathscr{T}$ and $X(x,y)=x$ and $Y(x,y)=y$, then $(x,y)\mapsto P[\cdot|Y])(x,y)$ depends only on $y$ (why?) Hence $y\mapsto f(P[\cdot|\sigma(Y)](y)]$ is $\mathscr{T}-\mathscr{B}(\mathbb{R})$ measurable.

A good reference for the basics of Random measures is Kallenberg, O. Foundations of Probability. 2nd edition, Springer 2001, p. 224-226.

For a comprehensive treatment random measures, stochastic geometry, foundations and applications can be found here

I'm a bit unsured about your notation, but the answer seems to be yes.

If $X$ is a Polish space (e.g. $\mathbb{R})$ then $P(X)$, the set of pronability measures, is again Polish. A (transition) kernel is a measurable mapping from some space $Y$ to $P(X)$.

The disintegration theorem states that if $\pi \in P(X \times Y)$ there is a kernel $(\pi^y)_y$ such that $\int f(x,y) d \pi = \int f(x,y) d\pi^y(x) dpr_Y\pi(y)$ for all measurable $f : X \times Y \to \mathbb{R}$.

In probabilistic language: If $\pi$ is the joint law of the $X$-valued RV $\xi$ and the $Y$-valued random variable $\eta$, then $pr_Y\pi$ the the law of $\eta$ (marginal distribution) and $\pi^y$ is the law of $\xi$ given $\eta=y$. (And the disintegration theorem is sometimes refferred as to "existence of a regular conditional distribution")

If you are given a measurable function $F: P(X) \to \mathbb{R}$, then cleary, $y \mapsto F(\pi^y)$ is measurable -- just as composition of two measuable functions, namely $F$ and the transition kernel.

Taking expectation is measurable (one the set, where it is well defined, which is a Borel subset of $P(X)$). This can be seen easily from the definition of the $\sigma$-algebra on $P(X)$.