I am trying to derive some equivalent of the transformation of variables formula. That is, given a random variable $Y=g(X)$, where $g$ is an invertible function, then the pdf of $Y$, $f_Y(y)$, is given by, $$ f_Y(y) = f_X(x)\left|\frac{dx}{dy}\right| \tag{1}. $$ This result is readily derived by considering the relevant cdf and applying the chain rule. I am trying to find some alternate derivation via marginalisation. The marginal distribution, $f_Y(y)$ can also be found by, \begin{align} f_Y(y) = &\int_\mathcal{X}f_{YX}(y,x)dx,\tag{2}\\ &=\int_\mathcal{X}f_{Y|X}(y|x)f_X(x)dx.\tag{3} \end{align}
Assuming that $g$ is invertible, I would have intuitively thought that $$ f_{Y|X}(y|x)=\delta(x-g^{-1}(y)),\tag{4} $$ where $\delta$ is the Dirac delta function. As such, $$ f_Y(y)=\int_\mathcal{X}\delta(x-g^{-1}(y))f_X(x)dx=f_X(g^{-1}(y)).\tag{5} $$ This is obviously incorrect. I believe I am making a mistake in (2), as the joint distribution is potentially not well defined? For context, I have a little knowledge of measure theory. Any help explaining where I made a mistake and how to correctly proceed would be greatly appreciated!
Their will be no joint distribution, because $Y$ is fully determined by $X$.
Rather this is basically a "change of variables" chain rule for integration.
$$\int_\mathcal Y f_Y(y)\,\mathrm dy = \int_\mathcal X f_X(h(y))\,\mathrm d h(y)$$
Where $h(y)=g^{-1}(y)$.