Let $F:\mathbb{R}^d \to [0,1]$ be an absolutely continuous joint cdf and let it be strictly increasing in each argument. Does it imply that its pdf $f$ is strictly positive a.s. (with respect to the Lebesgue measure on $\mathbb{R^d}$)? I would be very thankful for a reference to a book.
My idea was to show that $F$ being strictly increasing implies that $\int_A f d\lambda > 0$ for any $A = (a_1,b_1) \times \ldots \times (a_d, b_d)$, where $(a_i, b_i)$ are open intervals in $\mathbb{R}$ and $\lambda$ is the Lebesgue measure on $\mathbb{R}^d$. This would in turn imply that $f > 0$. However, I didn't manage to show any of these implications.
No, this is not true. We construct a counterexample in the plane. To make computations easy, we restrict ourselves to the unit square $U=[0,1]\times[0,1]$, but you will see that the idea extends to the entire plane easily. Define the set $N=[1/3,2/3]\times[1/3,2/3]$.
Consider the probability density function $f(x,y)=\begin{cases} 0&(x,y)\not\in U,\\ 0&(x,y)\in N,\\ 9/8&(x,y)\in U \setminus N. \end{cases}$
The associated marginal distribution functions are strictly increasing for arguments inside the unit interval. The idea easily extends to the whole space. For example, take a standard normally distributed vector in $\mathbb{R}^d$, set its density to zero for arguments in the unit ball and renormalize to get a density function again.