Is there an analogue Lebesgue decomposition theorem for joint distributions? Let $F:\mathbb{R}^n\rightarrow \mathbb{R}$ be a joint (cumulative) distribution function.
Then, can $F$ be expressed as a convex combination of absolutely continuous,discrete,singular continuous parts?
This is well-known to be true when $n=1$, but is this still true when $n>1$?
I think this is false because of “discrete” part. A joint distribution can be discontinuous at some point even if it is continuous coordinate-wisely, and this resists the above statement to be true. 
Is this true? If not, what kind of well-known decomposition is there for joint distributions?
 A: The Lebesgue decomposition theorem states (https://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem): 

For every two $\sigma$-finite signed measures $\mu$ and $\nu$ on a measurable space $(\Omega,\Sigma)$, there exist two $\sigma$-finite signed measures $\nu_0$ and $\nu_1$ such that:
  $$\nu =\nu_0+\nu_1,$$
  $\nu_0\ll \mu$ (that is, $\nu_0$ is absolutely continuous with    respect to $\mu$), 
$\nu_1\perp \mu$  (that is, $\nu_1$ and $\mu$ are    singular).

Let probability measure $P(B)=\mathbb P((X_1,\ldots,X_n)\in B)$ be the distribution of some random vector $(X_1,\ldots,X_n)$ on the Borel subsets of $\mathbb R^n$. It satisfies the Lebesgue decomposition theorem with respect to Lebesgue measure $\lambda$ which is restricted to Borel subsets of $\mathbb R^n$:
There exists two measures $\nu_0$ and $\nu_1$ such that: for every $B\in\mathfrak B(\mathbb R^n)$
$$ P(B)=\nu_0(B)+\nu_1(B).$$
Here $\nu_0\ll \lambda$, $\nu_1\perp \lambda$. Note also that both $\nu_0$ and $\nu_1$ are nonnegative finite measures such that
$$ P(\mathbb R^n)=\nu_0(\mathbb R^n)+\nu_1(\mathbb R^n)=1$$
and, excluding the trivial case when one of these measures is zero, one can define probability measures
$$P_0(B)=\frac{\nu_0(B)}{\nu_0(\mathbb R^n)}, \quad P_1(B)=\frac{\nu_1(B)}{\nu_1(\mathbb R^n)}$$ and recall the decomposition as
$$
P(B) = \alpha_0 P_0(B) + \alpha_1 P_1(B), \quad \alpha_0=\nu_0(\mathbb R^n), \alpha_1=\nu_1(\mathbb R^n).
$$
Here $P_0\ll \lambda$ is absolutely continuous with respect to Lebesgue measure. That is, there exists Radon-Nikodim derivative $f:\mathbb R^n\to \mathbb R$ such that for every Borel $B$
$$
P_0(B)=\int_B f \, d\lambda.
$$ 
Consider the singular part $P_1$. It can be decomposed further (look at the Wiki article cited above) as
$P_1=\nu_2+\nu_3$, where $\nu_2$ is a pure discrete measure and $\nu_3$ is a pure singular measure. That is, $P$ can be decomposed as
$$
P = \alpha_0 P_0 + \alpha_2 P_2 + \alpha_3 P_3, \quad \alpha_i\geq 0,\ \sum \alpha_i=1
$$
What is "pure" discrete measure? Probability measure $P_2$ is pure discrete w.r.t. $\lambda$ means that there exists at most countable set $S\subset \mathbb R^n$ s.t. $P_2(S)=1$. This measure is atomic: it is concentrated on the  set $S$.
And what is "pure" singular measure? Probability measure $P_3$ is "pure" singular w.r.t. $\lambda$ means that there exists a Borel set $B\subset \mathbb R^n$  s.t.: $\lambda(B)=0$, $P_2(B)=1$, and $P_2(\{x\})=0$ for any $x\in B$. 
What does it mean for CDF? Exactly the same: every CDF can be decomposed into three components
$$
F(x_1,\ldots,x_n)=P(\underline{(-\infty, x_1]\times\ldots\times(-\infty,x_n]}_B) = \alpha_0 P_0(B)+\alpha_2P_2(B)+\alpha_3P_3(B)=\alpha_0 F_0(x_1,\ldots,x_n)+\alpha_2F_2(x_1,\ldots,x_n)+\alpha_3F_3(x_1,\ldots,x_n).
$$
Here $F_0(x_1,\ldots,x_n)$ is absolutely continuous CDF, $F_2(x_1,\ldots,x_n)$ is the CDF of multivariate discrete distribution and $F_3(x_1,\ldots,x_n)$ corresponds to a singular distribution in $\mathbb R^n$, which is supported on some Borel subset of $\mathbb R^n$ with zero Lebesgue measure and does not have atoms. 
Note that in $\mathbb R^n$, $n\geq 2$, the third component can be discontinuous somewhere. Therefore, I deliberately do not call it a continuous singular component. Say, joint distribution of r.v.'s $X=1$, $Y\sim U(0,1)$ is singular in $\mathbb R^2$, and its joint CDF is discontinuous in every point of the set $\{(1,y): 0< y <1\}$. 
Thus, there is no difference in Lebesgue decomposition theorem in $\mathbb R$ and $\mathbb R^n$. The only difference is that in $\mathbb R$ the singular distribution is continuous, and in higher dimensions this is not necessarily true.
