Monotonicity with expectation

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).

Could you show if it holds or not?

• If you define $X(\omega) = Z_2(\omega) - Z_1(\omega)$ then $X(\omega)\geq 0$ for all $\omega \in \Omega$. You might use the Markov inequality on the random variable $X$. Commented Mar 12, 2020 at 18:30
• Are you familiar with the result that if $f \ge 0$ then $\int f d \mu = 0$ iff $f$ is zero ae. $[\mu]$? Commented Mar 12, 2020 at 18:39
• @Michael: I use that inequality and prove it. Commented Mar 16, 2020 at 20:20
• @Michael: I am wondering what if we assume $P(X(\omega)\geq 0)=1$ instead of $X\geq 0$? Commented Mar 23, 2020 at 14:46
• Sets of probability 0 do not affect expectations, so the result is the same. Equivalently you could define a new RV $Y=\max[X,0]$ which is always nonnegative, and $E[Y]=E[X]$. Commented Mar 23, 2020 at 16:35

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, $$\int_{\Omega_0}Z_1(\omega)\mathsf{P}(d\omega)<\int_{\Omega_0}Z_2(\omega)\mathsf{P}(d\omega).$$