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The idea here is to consider the convex set $F$ of measurable functions $f\colon\R^n\to[0,1]$ such that the one-dimensional marginals of $f(x)\,dx$ are $cp(t)\,dt$ for some real $c>0$. The set $F$ is nonempty, because the function $f$ given by $f(x_1,\dots,x_n)\equiv p(x_1)\cdots p(x_n)/\|p\|_\infty^n$ is in $F$. If now $F$ has an extreme point, it remains to show that any function in $F$ that is an extreme point of $F$ takes only values $1$ or $0$ almost everywhere (which looks very plausible, at least). Indeed, such an extreme point of $F$ coincides almost everywhere with the indicator of a measurable subset $A$ of $\R^n$ of nonzero Lebesgue measure such that each one-dimensional marginal of the uniform distribution $\U_A$ over $A$ is $\mu(dt)=p(t)\,dt$. This idea can of course be generalized to more general marginals.

The idea here is to consider the convex set $F$ of measurable functions $f\colon\R^n\to[0,1]$ such that the one-dimensional marginals of $f(x)\,dx$ are $cp(t)\,dt$ for $c:=1/\|p\|_\infty^n$. The set $F$ is nonempty, because the function $f$ given by $f(x_1,\dots,x_n)\equiv cp(x_1)\cdots p(x_n)$ is in $F$. If now $F$ has an extreme point, it remains to show that any function in $F$ that is an extreme point of $F$ takes only values $1$ or $0$ almost everywhere (which looks very plausible, at least). Indeed, such an extreme point of $F$ coincides almost everywhere with the indicator of a measurable subset $A$ of $\R^n$ of nonzero Lebesgue measure such that each one-dimensional marginal of the uniform distribution $\U_A$ over $A$ is $\mu(dt)=p(t)\,dt$. This idea can of course be generalized to more general marginals.

The idea here is to consider the convex set $F_C$ of measurable functions $f\colon\R^n\to[0,1]$ such that the one-dimensional marginals of $f(x)\,dx$ are $cp(t)\,dt$ for some real $c>0$. The set $F$ is nonempty, because the function $f$ given by $f(x_1,\dots,x_n)\equiv p(x_1)\cdots p(x_n)/\|p\|_\infty^n$ is in $F$. If now $F$ has an extreme point, it remains to show that any function in $F$ that is an extreme point of $F$ takes only values $1$ or $0$ almost everywhere (which looks very plausible, at least). Indeed, such an extreme point of $F$ coincides almost everywhere with the indicator of a measurable subset $A$ of $\R^n$ of nonzero Lebesgue measure such that each one-dimensional marginal of the uniform distribution $\U_A$ over $A$ is $\mu(dt)=p(t)\,dt$. This idea can of course be generalized to more general marginals.

The idea here is to consider the convex set $F$ of measurable functions $f\colon\R^n\to[0,1]$ such that the one-dimensional marginals of $f(x)\,dx$ are $cp(t)\,dt$ for some real $c>0$. The set $F$ is nonempty, because the function $f$ given by $f(x_1,\dots,x_n)\equiv p(x_1)\cdots p(x_n)/\|p\|_\infty^n$ is in $F$. If now $F$ has an extreme point, it remains to show that any function in $F$ that is an extreme point of $F$ takes only values $1$ or $0$ almost everywhere (which looks very plausible, at least). Indeed, such an extreme point of $F$ coincides almost everywhere with the indicator of a measurable subset $A$ of $\R^n$ of nonzero Lebesgue measure such that each one-dimensional marginal of the uniform distribution $\U_A$ over $A$ is $\mu(dt)=p(t)\,dt$. This idea can of course be generalized to more general marginals.

The idea here is to consider the convex set $F_C$ of measurable functions $f\colon\R^n\to[0,1]$ such that the one-dimensional marginals of $f(x)\,dx$ are $\mu(dt)=cp(t)\,dt$ for some real $c>0$. The set $F$ is nonempty, because the function $f$ given by $f(x_1,\dots,x_n)\equiv p(x_1)\cdots p(x_n)/\|p\|_\infty^n$ is in $F$. If now $F$ has an extreme point, it remains to show that any function in $F$ that is an extreme point of $F$ takes only values $1$ or $0$ almost everywhere, which looks very plausible, at least. Indeed, such an extreme point of $F$ coincides almost everywhere with the indicator of a measurable subset $A$ of $\R^n$ of nonzero Lebesgue measure such that each one-dimensional marginal of the uniform distribution $\U_A$ over $A$ is $\mu(dt)=cp(t)\,dt$. This idea can of course be generalized to more general marginals.

However, this extreme-point construction is not explicit and the resulting set may turn out to have poor topological properties. In this respect, the simple and explicit construction (1) may still be of interest, even though it only approximates the marginals and works only when all the one-dimensional marginals are equal to one another. $\Box$

The idea here is to consider the convex set $F_C$ of measurable functions $f\colon\R^n\to[0,1]$ such that the one-dimensional marginals of $f(x)\,dx$ are $cp(t)\,dt$ for some real $c>0$. The set $F$ is nonempty, because the function $f$ given by $f(x_1,\dots,x_n)\equiv p(x_1)\cdots p(x_n)/\|p\|_\infty^n$ is in $F$. If now $F$ has an extreme point, it remains to show that any function in $F$ that is an extreme point of $F$ takes only values $1$ or $0$ almost everywhere (which looks very plausible, at least). Indeed, such an extreme point of $F$ coincides almost everywhere with the indicator of a measurable subset $A$ of $\R^n$ of nonzero Lebesgue measure such that each one-dimensional marginal of the uniform distribution $\U_A$ over $A$ is $\mu(dt)=p(t)\,dt$. This idea can of course be generalized to more general marginals.

However, this extreme-point construction is not explicit and the resulting set $A$ may turn out to have poor topological properties. In this respect, the simple and explicit construction (1) may still be of interest, even though it only approximates the marginals and works only when all the one-dimensional marginals are equal to one another. $\Box$