Let $X \in L_1$ be an integrable random variable and $\mathcal{F}$ a $\sigma$-algebra.
Suppose
$$\mathbb{E}[X|\mathcal{F}]\geq 0.$$
Let $Z\geq 0 $ be a non-negative ranfom variable. Do we have
$$\mathbb{E}[ZX|\mathcal{F}]\geq 0?$$
My idea is to show for any $A \in \mathcal{F}$, $$\int_{A}\mathbb E[ZX|\mathcal{F}]dP = \int_{A}ZXdP\geq 0$$ using the fact that $$\int_{A}\mathbb E[X|\mathcal{F}]dP = \int_{A}XdP\geq 0.$$ But it is hard for me to make a rigorous argument.
Let $\Omega=\{1,2,3,4\}$, $\mathcal F=\{\emptyset,\{1,2\},\{3,4\},\Omega\}$.
Let $$X(1)=-1,X(2)=1,X(3)=0,X(4)=0$$ and
$$Z(1)=100,Z(2)=Z(3)=Z(4)=0.$$
Finally let $p_1=P(\{1\})=\frac1{10},\ p_2=P(\{2\})=\frac2{10}, \ p_3=P(\{3\})=\frac2{10}, \ p_4= P(\{4\})=\frac5{10}$.
Now,
$$E[X\mid \mathcal F](\omega)=\begin{cases}\frac{p_2-p_1}{p_1+p_2}=\frac13&\text{ if } &\omega\in A\\ 0&\text{ if } &\omega\in B. \end{cases}$$
At the same time
$$E[ZX\mid \mathcal F](\omega)=\begin{cases}-100\frac{p_1}{p_1+p_2}=-\frac{100}3&\text{ if } &\omega\in A\\ \ \ \ \ 0&\text{ if } &\omega\in B. \end{cases}.$$
That is, $Z$ is able to attenuate the negative part of $X$.