Conditional expectation given event from first principles

Question

Given an event $H$, and a random variable $X$ defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, the following two are given usually as a definition: either $\mathbb{E}(X|H)=\mathbb{E}(X1_H)/\mathbb{P}(H)$, or $\mathbb{E}(X|H)=\int_{\Omega} Xd\mathbb{P}_H(\omega)$, where $\mathbb{P}_H(A)=\mathbb{P}(A\cap H)/\mathbb{P}(H)$ for every event $A$. Clearly one should be able to derive the former from the latter (which, to me, seems more fundamental). I have tried to perform the derivation, which I was expecting to be trivial, but I stumbled on a measure theoretic detail. Since $\mathbb{P}(H^c|H)=0$, $\int_{\Omega} Xd\mathbb{P}_H(\omega)=\int_{H^c} Xd\mathbb{P}_H(\omega)+\int_{H} Xd\mathbb{P}_H(\omega)=\int_{H} Xd\mathbb{P}_H(\omega)=\int_{\Omega} X1_Hd\mathbb{P}_H(\omega)$. Here I do not know how to go rigorously from $d\mathbb{P}_H(\omega)$ to $d\mathbb{P}(\omega)/\mathbb{P}(H)$, which is clearly the last step that yields the result sought. Thank you for any insight you might be able to provide. EDIT: I forgot to mention, that I can actually prove the identity, but not with elementary methods. In particular by using the sigma algebra $\mathcal{H}$ generated by $H$ and $\mathbb{E}_{\mathcal{H}}X=1_H\mathbb{E}(X1_H)/\mathbb{P}(H)+1_{H^c}\mathbb{E}(X1_{H^c})/\mathbb{P}(H^c)$ and the consequent conditional measure $\mathbb{P}_{\mathcal{H}}^\omega(\cdot)=1_H(\omega)\mathbb{P}(\cdot|H)+1_{H^c}(\omega)\mathbb{P}(\cdot|H^c)$, exploiting the fact that $(\mathbb{E}_{\mathcal{H}}X)(\omega)=\int_{\Omega} X\mathbb{P}_\mathcal{H}^\omega(d\omega)$, choosing any $\omega\in H$ yields the desired result. However, I am looking for a proof relying only on elementary conditioning and distributions, which does not involve conditioning on sigma algebras. It feels like such a proof should be possible.

Answer 1

I actually ended up figuring it out for myself. I write the answer as it may be of use to others. The more elementary approach is to exploit Radon-Nikodym theorem, due to the absolute continuity of $\mathbb{P}_H$ with respect to $\mathbb{P}$, and show, quite trivially, that the derivative $d\mathbb{P}_H/d\mathbb{P} = 1_H/\mathbb{P}(H)$, through the relationship $\mathbb{P}(A|H)=\int_A 1_H/\mathbb{P}(H)d\mathbb{P}$.