If $E(X \mid A)\ge c$ and $E(X \mid B)\ge c$, then $E(X \mid A\cup B)\ge c$

Question

Let $A$ and $B$ be two disjoint events and $X$ any random variable with finite expectation. I have to prove that if $E(X \mid A)\ge c$ and $E(X \mid B)\ge c$, then $E(X \mid A\cup B)\ge c$.

We have

$$E(X \mid A\cup B)=\frac{\int_A X dP + \int_B X dP}{P(A)+P(B)}$$ but how can I go from here?

Answer 1

Hint: Use $$\frac{\int_A X \, d\mathbb{P}}{\mathbb{P}(A)+\mathbb{P}(B)} = \frac{\mathbb{P}(A)}{\mathbb{P}(A)+\mathbb{P}(B)} \underbrace{\frac{\int_A X \, d\mathbb{P}}{\mathbb{P}(A)}}_{\geq c} \geq c \frac{\mathbb{P}(A)}{\mathbb{P}(A) +\mathbb{P}(B)}$$ and the corresponding estimate for $\frac{\int_B X \, d\mathbb{P}}{\mathbb{P}(A)+\mathbb{P}(B)}$.

Answer 2

Note that $$\int_A X\mathrm d\mathbb P\geqslant c\mathbb P(A);$$ the same bound holds when $A$ is replaced by $B$. Then take the sum.