Singular Radon probabilities on $[0,1]^d$. Is conditioning on $x_i = \alpha$ well-defined?

Question

The question: Let $\pi$ be a Radon probability measure on $[0,1]^d$, $2\leq d < \omega$, that is singular (w.r.t. to the $d$-dimensional Lebesgue measure). Suppose that for $i\in \{1,\dots,d\}$ and $\alpha \in [0,1]$, the set $S = \{ x \in [0,1]^2 \mid x_i = \alpha \}$ intersects the support of $\pi$ (but might have $\pi(S)=0$).

Is there some canonical way of defining probabilities conditional on $S$? If so, what is it, and are my assumptions sufficient to make it well-defined?

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Informal motivatation: I have a probability distribution on $\pi$ on $[0,1]^d$, $d\geq 2$ (possibly $d=\omega$, but let's keep it finite for the sake of the question). I need to know which additional assumptions to make s.t. the conditional probabilities $\mathbf P_\pi [\cdot \mid x_i=\alpha]$ for $i\leq d$, $\alpha \in [0,1]$, are well-defined.

My observations: This is, obviously, fine if $\pi ( \{x\in [0,1]^d \mid x_i = \alpha\} ) > 0$. More generally, I can assume that $\pi$ is a Radon probability measure, decompose it into the absolutely continuous part and singular part as $\pi = \pi_a + \pi_s$, and assume that the section $x_i=\alpha$ intersects the support. Dealing with the $\pi_a$ part is straightforward. However, I don't know what to do with $\pi_s$.

Answer 1

$\newcommand{\B}{\mathcal B}$ First here, $[0,1]^d$ is a Polish space (i.e., a separable complete metric space). So, $[0,1]^d$ is a Radon space, and hence any (Borel) probability measure is Radon. So, you did not have to say that the probability measure $\pi$ on $[0,1]^d$ is Radon. (Also, you did not have to assume that $\pi$ is singular.)

Anyway, it follows that for each $i=1,\dots, d$ your probability measure $\pi$ admits a regular conditional probability distribution $[0,1]\times\B_d\ni(t,A)\mapsto\nu_i(t,A)\in[0,1]$, such that the map $[0,1]\ni t\mapsto\nu_i(t,A)\in[0,1]$ is $\B_1$-measurable for each $A\in\B_d$, the map $\B_d\ni A\mapsto\nu_i(t,A)\in[0,1]$ is a probability measure for each $t\in[0,1]$, and $$\pi(A\cap p_i^{-1}(B))=\int_B\nu_i(t,A)\pi(p_i^{-1}(dt)) $$ for all $A\in\B_d$ and $B\in\B_1$, where $\B_k$ is the Borel $\sigma$-algebra over $[0,1]^k$ and $[0,1]^d\ni x=(x_1,\dots,x_d)\mapsto p_i(x):=x_i\in[0,1]$.

So, you can let $\mathbf P_\pi [\cdot \mid x_i=\alpha]:=\nu_i(\alpha,\cdot)$ for all $\alpha\in[0,1]$.

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One may note here that a regular conditional probability distribution $\nu_i$ is of course not (quite) uniquely defined. For instance, for any $N\in\B_1$ such that $\pi(p_i^{-1}(N))=0$ and all $t\in N$, one can replace $\nu(t,\cdot)$ by any probability measure $\rho$ on $\B_d$.