Suppose we have a random vector $\mathbf{X} = (X_1,\ldots,X_n)$ or sample with pdf or pmf $f(\mathbf{x};\theta)$, where $\theta \in \Theta$ is an unknown parameter (or vector of parameters), whose specification completely determines $f(\mathbf{x};\theta)$.
As far as I understand, a statistic is just a a random variable $T=T(\mathbf{x})$ that is a function (that does not depend on $\theta$) of $\mathbf{X}$. That's not to say the distribution of $T$ doesn't depend on $\theta$ of course, just that the function that relates $\mathbf{X}$ and $T$ does not.
Now, usually from what I've seen, the statistic $T$ is defined to be a sufficient statistic if the distribution of $\mathbf{X}$ conditional on $T=t$ does not depend on $\theta$.
Now, my issue with this definition (which I don't understand how is it not addressed anywhere I've looked) is that if the support of $T$ depends on $\theta$, then we can't really pick a $t$ in the support of $T$ and then claim that the conditional distribution does not depend on $\theta$, because that same conditional distribution may be ill-defined for some values of $\theta$ for which $t$ is not in the support of $T$, right?
I kinda came up with my own definition which I'm hoping someone could comment on to see if this is what's more formally meant by that usual definition:
A statistic $T(\mathbf{X})$ is sufficient for $\theta$ if there exists a non-negative function $C:\mathbb{R}^n\times \mathbb{R} \to \mathbb{R}$ that does not depend on $\theta$ and such that $$p(\mathbf{x},t;\theta) = C(\mathbf{x},t) \cdot q(t;\theta)$$ for all $\mathbf{x} \in \mathbb{R}^n, t\in \mathbb{R}, \theta \in \Theta$, where $p$ and $q$ denote the joint pdf/pmf of $(\mathbf{X},T(\mathbf{X}))$ and the pdf/pmf of $T(\mathbf{X})$, respectively.