I'm learning about the proof of Fisher's factorization theorem online. Almost all the proofs of it in case of continuous is using the one-to-one transformation. But I can't find the necessary of transformation.
I assume that $f_{\theta}(x)=h(x)g(T(x);\theta)$,then I want to prove $T(x)$ is sufficient for $\theta$
I think the proof:
$f_{\theta}(x|t)=\frac{f_{\theta}(x,t)}{f_{\theta}(t)}\\ =\frac{f_{\theta}(x)}{\int\ldots\int_{x;T(x)=t}f_{\theta}(x,t)dx}\\=\frac{h(x)g(t;\theta)}{\int \ldots \int_{x;T(x)=t} h(x)g(t;\theta)dx}\\=\frac{h(x)g(t;\theta)}{g(t;\theta)\int \ldots \int_{x;T(x)=t}h(x)dx}\\=\frac{h(x)}{\int\ldots\int_{x;T(x)=t}h(x)dx}$
The last formula is independent of $\theta$, so the conditional pdf given T(x)=t is sufficient statistics.
Is there anything wrong with what I wrote ?
Thank you for helping me and please bear with my poor English!