I have a set of data D for which I know it's distribution, given 3 parameters A, B, C. What I want is the posterior multidimensional distribution P(A, B, C | D). I already have the priors P(A), P(B) and P(C).
Starting with:
$P(A,\ B,\ C\ |\ D) = \cfrac{P(A,\ B,\ C,\ D)}{P(D)}=\cfrac{P(D,\ A,\ B,\ C)}{P(D)}$
and using the law of total probability, I get:
$P(A,\ B,\ C\ |\ D) = \cfrac{P(D\ |\ A,\ B,\ C)\ P(A\ |\ B,\ C)\ P(B\ |\ C)\ P(C)}{P(D)}$
Assuming A, B and C are independent, then $P(A\ |\ B,\ C)\ =\ P(A)$ and $P(B\ |\ C)\ =\ P(B)$, right?
And for P(D), can I write it like this?
$P(D)\ =\ \int P(D|A)\ P(A)dA\ =\ \int\int P(D|A,\ B)\ P(A)\ P(B)\ dA\ dB \\ \ \ \ \ \ \ \ \ \ =\ \int\int\int P(D|A,\ B)\ P(A)\ P(B)\ P(C)\ dA\ dB\ dC$
Therefore, my posterior would be
$P(A, B, C | D) = \cfrac{P(D | A, B, C)\ P_0(A)\ P_0(B)\ P_0(C)}{\int\int\int{P(D | A, B, C)\ P_0(A)\ P_0(B)\ P_0(C) \ dA\ dB\ dC}}$
where $P_0(A)$, $P_0(B)$, $P_0(C)$ are my priors.
Am I missing something? Is this argument wrong? How would one proceed if A, B and C are not independent?
Indeed, your argument is valid. Using your generalised notation, Bayes' rule and the Law of Total Probability do say:
$$\begin{align}P(A,B,C\mid D) ~&=~ \dfrac{P(D\mid A,B,C)~P_0(A,B,C)}{P(D)}\\[1ex] ~&=~ \dfrac{P(D\mid A,B,C)~P_0(A,B,C)}{\iiint P(D\mid A,B,C)~P_0(A,B,C)~\mathsf d C~\mathsf d B~\mathsf d A}\end{align}$$
Mutual independence of $A,B,C$ would then obtain your final result. If you do not have such independence, then you simply need the joint probability measure for $A,B,C$; or the relevant conditional measures.