Consider the probability space $(\Omega, \mathcal{A},\mathbb{P})$, where $\Omega = (-1,1), \mathcal{A} = \mathcal{B((-1,1))}$ and $\mathbb{P}$ is the uniform distribution on $(-1,1)$. For an integrable RV $X$ compute the conditional expectation $\mathbb{E}(X \mid \mathcal{F})$, where $\mathcal{F} = \{A \in \mathcal{A} : A = -A \}$.
Remark: As usual $\mathcal{B}(\cdot)$ denotes the Borel Algebra.
I know that for $\mathbb{E}(X \mid \mathcal{F})$ is uniquely defined to via the conditions $\mathcal{F}$-measurable, $\mathbb{E}(X \mid \mathcal{F}) \in L^1(\mathbb{P})$ and $\int_A \mathbb{E}(X \mid \mathcal{F}) d \mathbb{P} = \int_A X d \mathbb{P}$.
So, since $\mathcal{F}$ consists of symmetric intervals around $0$ I suppose (intuitively) that $\mathbb{E}(X \mid \mathcal{F}) = 0$.
However, I don't know how to formally argue (I am not good at measure theory) that in our case we have
$$\int_A X d \mathbb{P} = 0 \quad (\forall A \in \mathcal{A}).$$
Could you please give me a hint?
Edit: Thanks to LucaMac's comment I realise that my idea was flawed, but what about $\mathbb{E}(X \mid \mathcal{F}) = X(0)$?