I am interested in finding the extreme points of the following set of distributions 
\begin{align}
\mathcal{P}= \left\{F: \int\_{\mathbb{R}} |x|^k dF(x)=c \right\}
\end{align}
where $k,c>0$. 


I know that [this paper](https://www.jstor.org/stable/pdf/3689944.pdf?refreqid=excelsior%3A6c1f2a909f95192f8b30edf1e31ad3de) by Winkler is a standard reference on this question. 


I was trying to extract the answer from this paper and got that the set of extreme points is given by 
\begin{align}
\mathop{\rm ex}(\mathcal{P}) = \left\{F \in \mathcal{P} : F= (1-t) \delta\_{x\_1}+t \delta\_{x\_2}, t\in[0,1], x\_1,x\_2 \in \mathbb{R} \right\}.
\end{align}


However, not sure if this is correct since in [this question](https://mathoverflow.net/questions/85527/extreme-points-of-a-set-of-probability-measures), for a very similar set, it was pointed out that it must be combination of three mass points instead of two. 



Extreme points of set of probability measures $\mathcal{P}= \{F: \int_{\mathbb{R}} |x|^k dF(x)=c \}$
What the paper asserts in this case is $ex\mathcal P=\{F:F=(1-t)\delta\_{x}+t\delta\_y, t\in[0,1], (1-t)|x|^k+t|y|^k=c, x+y\neq0\}$. This includes singletons $\delta\_x$ with $|x|^k=c$.


What was pointed out is not the existence of three-point masses as extreme points, but that the proof (that only convex combinations of two point masses are extreme) was obvious for probability measures on a three point set instead of $\mathbb R$.


