According to
https://en.m.wikipedia.org/wiki/Conditional_expectation, it is not possible to compute integrals of the form;
$\int X dP_{H}$.
Where $dP_{H}$ is the restriction of $P$.
I have a hard time seeing where the shoe does not fit. The "restriction" should have a value for each set since it has a value in some superset.
Is there a problem because X might take values in some Borel set which does not have well defined preimage or is the problem that we might divide by zero. If the latter is the case then the measure $dP_{H}$ is not only a restriction but a rescaled restriction.
Your final paragraph is on the right track. The problem is the pre-image of an open set under $X$ may not be in $H$. That is there might be an open set $U$ such that $$X^{-1}(U)\notin H.$$ In this case we say that $X$ is not $H$-measurable, and $\int X\,\mathrm{d}\mathbf{P}_H$ is not defined since the Lebesgue integral is only defined for measurable functions.
To give some intuition for this, suppose $F$ is a measurable set which is not in $H$, and $X=\mathbf{1}_F$ is the indicator of $F$. From the way the Lebesgue integral is defined: $$\int X\,\mathrm{d}\mathbf{P}=\int \mathbf{1}_F\,\mathrm{d}\mathbf{P}=\mathbf{P}(F)$$ But since $F\notin H$, $\mathbf{P}_H(F)$ is not defined, and therefore neither is $\int X\,\mathrm{d}\mathbf{P}_H$.