This is a follow-up to my previous question:
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 $(-1,1)$. Let further $\mathcal{F} = \{A \in \mathcal{A} : A = -A \}$. For an integrable RV $X$ show that $$\int_A\dfrac 12 (X(|\omega|) + X(-|\omega|)) d \mathbb{P} = \int_A X d \mathbb{P}$$ for all $A \in \mathcal{F}$.
Remark: As usual $\mathcal{B}(\cdot)$ denotes the Borel Algebra.
I understand that the $A$ describe symmetric intervals around $0$. Using the linearity of the integral we may write
$$\int_A \dfrac 12 (X(|\omega|) + X(-|\omega|)) d \mathbb{P} = \dfrac 12 \int_A (X(|\omega|) d \mathbb{P}+ \dfrac 12 \int_A X(-|\omega|)) d \mathbb{P}$$
, but I fail to see how to now compute the summands explicitly. (I am not good in measure theory.) Could you please give me a hint?
$$ \int_A X d\mathbb P = \frac12 \int_A X(t)\;dt,\qquad\text{(Lebesgue integral).} $$ Let $B = A \cap (0,1)$. Then, because $A$ is symmetric, we have $$ A = B \cup (-B)\qquad\text{a disjoint union} $$ except perhaps for the point $\{0\}$ which has measure zero. So $$ \frac12\int_A X(t)\;dt = \frac12\int_B X(t)\;dt + \frac12 \int_{-B}X(t)\;dt $$ change variables $t=-s$ in the second one: $$ \frac12\int_B X(t)\;dt + \frac12\int_B X(-s)\;ds $$ and rename a dummy variable $$ \frac12 \int_A X(t)\;dt = \frac12\int_B X(t)\;dt + \frac12\int_B X(-t)\;dt \tag1$$ This is probably the best way to write the answer.
But the OP has written it in a more complicated way. How can we get that?
Similar to our derivation of $(1)$ [change variables in the first term, not the second], we may obtain $$ \frac12 \int_A X(t)\;dt = \frac12 \int_{-B} X(-t)\;dt + \frac12 \int_{-B} X(t)\;dt \tag2$$ Add $(1)$ and $(2)$ \begin{align} \int_A X(t)\;dt &= \frac12\int_B X(t)\;dt + \frac12\int_B X(-t)\;dt + \frac12 \int_{-B} X(-t)\;dt + \frac12 \int_{-B} X(t)\;dt \\ & = \frac12\int_{B \cup (-B)} X(t)\;dt + \frac12\int_{B\cup(-B)} X(-t)\;dt \\ & = \int_A\frac12\big(X(t)+X(-t)\big)\;dt \end{align} the two values $t$, $-t$ are the same as the two values $|t|, -|t|$ in some order, so $$ \int_A X(t)\;dt = \int_A\frac12\big(X(|t|)+X(-|t|)\big)\;dt \\ \frac12\int_A X(t)\;dt = \frac12\int_A\frac12\big(X(|t|)+X(-|t|)\big)\;dt $$ and finally $$ \int_A X(\omega)\;d\mathbb P = \int_A \frac12\big(X(|\omega|)+X(-|\omega|)\big)\;d\mathbb P \tag3$$ Why would you want to write $(1)$ in this more complicated form $(3)$? Because now we can apply the definition of conditional expectation and conclude $$ \mathbb E [ X \mid \mathcal F ](\omega) = \frac12\big(X(|\omega|)+X(-|\omega|)\big)\qquad\text{a.s.} \tag4$$