@losifPinelis , could you cite some reference for the expressions, e.g. (1) ?

My intuition is that the image measure $Q \zeta $ can be written as the Lebesgue measure in $\mathbb R^{k}$. Ok. I agree, working with the spectral decomposition, makes everything nicer, because you are working with $f_2 = f_1 \circ Q^\intercal = f_{\sigma_1} * \dots * f_{\sigma_k}$ (of course $f_{\sigma_i}$ consider only the components of the Eigenvectors that don't have EV 0). But, I would like to work with $f_1$ and $\zeta$ (if possible).

I wish to write $$ Pr(X \in A) = \int_A f_1(x) d\zeta \quad (A1)$$ where $$ f_1(x) = \frac{1}{2 \pi^{c-1} |\Sigma|} e^{-\frac{1}{2} x^\intercal \Sigma^- x}. $$ I wish to work with $f_1$, and wonder how the measure $\zeta$ looks like. If you apply a transformation as you mention above $Y = Q^\intercal X$, then we are just applying a transformation formula $$ \int_{A} f_1 Q^\intercal Q d\zeta = \int_{Q A} f_2(x) d Q(\zeta)$$ and $f_2= f_{\sigma_1} * \dots f_{\sigma_k}$ (with your not. as in (1) )(since the cov. matrix is now diagonal) $Q \zeta=$ image measure of $\zeta$ under $Q$ (as oper.)

thanks again! Sadly I feel, that you don't get that my question is about, how to write the degenerate measure in a close form (if possible), because I want to keep the "density function". I guess this degenerate measure cannot be the image measure of the Lebesgue measure under a transformation. Please consider my question again (last time): I have $X \sim \mathcal N(0, \Sigma)$. Sigma is only positiv semi-definit. I consider the generalized inverse $\Sigma^-$ and generalized determinant $|\Sigma|$ (which cannot (!) be zero), because I ignore the 0 eigenvalues.

@losifPinelis: Thanks a lot for the effort!! I understand what you mean and that your explanation help for numerical computations. But my question was slightly different as your answer. I formulate again with the same terminology: Given is a positiv-SEMI definite matrix $\Sigma$, $\mu = 0$ and $$ \pi(x) = \frac{1}{2 \pi^{c-1} |\Sigma| } e^{-\frac{1}{2} x^T \Sigma^- x},$$ $|\Sigma|, \Sigma^-$ are generalized determinant and inverse. Is there a way to write the degenerate measure $\zeta$ in a close form, i.e. if $$Pr(X \in B) = \int_B \pi(x) d\zeta,$$ then $$\zeta = ?$$

Thanks for your answer! This is what I guessed, when I meant using $\widetilde \pi$, since $$ \widetilde \pi(y) = f_{\sigma 1} (y_1) \cdots f_{\sigma k}(y_k). $$ Thus, $\widetilde \pi$ can be seen as the density function of $Q^\intercal X \sim \mathcal N (0,D)$ "in a sense" (avoiding the zero components). But, is there a way to express the density of $X$ (i.e., $\pi$ in my question) w.r.t. a (degenerate) measure?, i.e., $$P(X\in B) = \int \pi(x) d?, $$ The reason for this need, is that I want to apply later transformations, and applying already one, makes everything more confusing.