Proving $X$ is $\mathcal{A}^{\prime}$-measurable if and only if $Y$ is $\mathcal{A}^{\prime}$-measurable, provided that $X=Y$ a.s. Let $(\Omega, \mathcal{A}, P)$ be a probability space and $\mathcal{A}^{\prime}$ be the $P$-completion of $\mathcal{A}$. Let $X,Y$ be real-valued random variables defined on $\mathcal{A}^{\prime}$. I need to show that $X$ is $\mathcal{A}^{\prime}$-measurable if and only if $Y$ is $\mathcal{A}^{\prime}$-measurable.
I have looked up for any hints on this website or google and it seems that you can decompose the set $X^{-1}(A)$ into
$$(Y^{-1}(A) \cap (Y^{-1}(A) \cap X^{-1}(A)^c)^c) \cup (X^{-1}(A) \cap Y^{-1}(A)^c)$$
However, I cannot find any explanation of why this is true. Also, what is the set $A$? Is it an arbitrary subset of $\mathbb{R}$? How come that $X^{-1}(A) \cap Y^{-1}(A)^c$ is a null set? I know that $\{\omega : X(\omega) \neq Y(\omega)\}$ is a null set by the assumption. 
 A: There is a typo in the second sentence of your question. What's stated is a tautology. I will show the result stated in the title in the manner you indicated. 
First, $A$ is standardly assumed to be a Borel subset of $\mathbb R$ (though it doesn't really matter what sigma-algebra we eqip $\mathbb R$ with for the purposes of this result). To show that $X$ is $\mathcal A'$-measurable, we must show that $X^{-1}(A) \in \mathcal A'$.
To get the identity you ask about, start by noting that in general for any set $\Omega$ and any subsets $E,F$ of $\Omega$, the following hold:
$$E = (E \cap F) \cup (E \cap F^c) \tag{1}$$
and
$$F \cap (F \cap E^c)^c = F \cap (F^c \cup E) = \emptyset \cup (E \cap F) = E \cap F.\tag{2}$$
(In (2), we use DeMorgan's law and Distribution, which you can Google if you're unfamiliar. You can verify (1) with a Ven diagram.)
Now use (1) and (2) with $E = X^{-1}(A)$ and $F = Y^{-1}(A)$ to get 
$$X^{-1}(A) = (Y^{-1}(A) \cap (Y^{-1}(A)) \cap X^{-1}(A)^c)^c \cup (X^{-1}(A) \cap Y^{-1}(A)^c). \tag{3}$$
To conclude, we show that the set on the right-hand side of (3) is in $\mathcal A'$. Start by observing that if $\omega \in X^{-1}(A) \cap Y^{-1}(A)^c$, then $X(\omega) \in A$ and $Y(\omega) \in A^c$, which implies that $X(\omega) \neq Y(\omega)$. Thus,
$$X^{-1}(A) \cap Y^{-1}(A)^c \subseteq \{\omega: X(\omega) \neq Y(\omega)\}.$$
Since $P(X \neq Y)=0$ and $\mathcal A'$ is the $P$-completion of $\mathcal A$,
$$X^{-1}(A) \cap Y^{-1}(A)^c \in \mathcal A'.\tag{4}$$ 
Since $\mathcal A'$ is a sigma-algebra, 
$$(X^{-1}(A) \cap Y^{-1}(A)^c)^c \in \mathcal A'.\tag{5}$$
Since $Y$ is assumed to be $\mathcal A'$-measurable, $Y^{-1}(A) \in \mathcal A'$, and then by (5):
$$Y^{-1}(A) \cap (X^{-1}(A) \cap Y^{-1}(A)^c)^c \in \mathcal A'.\tag{6}$$
By (3), (4), and (6), we conclude that $X^{-1}(A) \in \mathcal A'$, as desired.
