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Given $X_{1,2,\cdots,d}:(\Omega,\mathcal F,\Bbb P)\to \Bbb R$ are random variables with absolutely continuous distributions $F_i(x_i)$. Consider their joint "CDF": $$F(x_1,\cdots,x_d):=\Bbb P(X_1\le x_1,\cdots,X_d\le x_d).$$ Then $F$ induces a probability measure on $\Bbb R^d$. Now my question is, how to prove that there exists a unique non-negative, measurable function $f:\Bbb R^d\to\Bbb R_{\ge 0}$ such that:
1). $$f(x_1,\cdots,x_d)=p(x_1,\cdots,x_d)=\dfrac{\partial^d F(x_1,\cdots,x_d)}{\partial x_1\cdots\partial x_d}$$ almost everywhere;
2). $$ F(x_1,\cdots,x_d)=\int_{(-\infty,x_1]\times\dots\times(-\infty,x_d]}f(x_1,\cdots,x_d)\,\mathrm dx_1\cdots\mathrm d{x_d}.$$

I think once we prove the integration of $p$ and that of $f$ coincides on rectangles then uniqueness follows. So actually we only have to show that $\int p = F$ where the integration is over $(-\infty, x_1]\times\cdots\times(-\infty,x_d]$. Looks easy but I still have clue. Any suggestion? Thanks!

EDIT As per NCh's comment the partial derivative may not exist even when each component's CDF is absolutely continuous. So let's just assume this singularity away, i.e., the partial derivative exists a.e.

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    If the marginals are absolutely continuous, joint density does not necessarily exist. Say, for vector $(X,X)$ with absolutely continuous distribution of $X$ the distribution is concentrated on the line $x_1=x_2$ in $\mathbb R^2$. And partial derivatives are equal to 0 a.e. The density does not exist. – 2017-02-12
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    @NCh sorry, my bad. Let's then suppose the partial derivative exists in the first place. (I don't know what's the proper condition for this, though. In one-dimensional cases it is equivalent to absolute continuity; now in $\Bbb R^d$ I'm not sure if it's also equivalent to the condition $\Bbb P\ll \mu$ where $\mu$ is the Lebesgue measure. – 2017-02-12

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