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$\newcommand{\B}{\mathcal B}$ 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 $\alpha\in[0,1]$.

One may note here 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 $\B_d$.

$\newcommand{\B}{\mathcal B}$ 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 $\alpha\in[0,1]$.

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 $\B_d$.

$\newcommand{\B}{\mathcal B}$ 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 $\alpha\in[0,1]$.

$\newcommand{\B}{\mathcal B}$ 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 $\alpha\in[0,1]$.

One may note here 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 $\B_d$.

$\newcommand{\B}{\mathcal B}$ 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 $[0,1]^d\ni(x_1,\dots,x_d)\mapsto p_i(x):=x_i\in[0,1]$.

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

$\newcommand{\B}{\mathcal B}$ 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 $\alpha\in[0,1]$.