I think the following is true but I cannot prove it.
Let $Z_1, Z_2$ are two random variables defined on the same sample space $\Omega$. Suppose that $Z_1(\omega) < Z_2(\omega)$ for all $\omega\in \Omega_0$ and $Z_1(\omega) = Z_2(\omega)$ for all $\omega\in \Omega\setminus\Omega_0$. We have:
If $P(\Omega_0)=0$, i.e. $Z_1=Z_2$ almost surely, then $E(Z_1)=E(Z_2)$.
If $P(\Omega_0)>0$, then $E(Z_1)<E(Z_2)$.
Could you show if it holds or not?
Write $$ \mathsf{E}Z_1=\int_{\Omega}Z_1(\omega)\mathsf{P}(d\omega)=\int_{\Omega_0}Z_1(\omega)\mathsf{P}(d\omega)+\int_{\Omega\setminus\Omega_0}Z_1(\omega)\mathsf{P}(d\omega).\tag{1}\label{1} $$ Since $$ \int_{\Omega\setminus\Omega_0}Z_1(\omega)\mathsf{P}(d\omega)=\int_{\Omega\setminus\Omega_0}Z_2(\omega)\mathsf{P}(d\omega), $$ the result depends on the probability of $\Omega_0$. If $\mathsf{P}(\Omega_0)=0$, then the first integral on the RHS of $\eqref{1}$ is $0$ (see, e.g., this question). Otherwise, $$ \int_{\Omega_0}Z_1(\omega)\mathsf{P}(d\omega)<\int_{\Omega_0}Z_2(\omega)\mathsf{P}(d\omega). $$