what does this set definition mean defined on independent random variables? Let $X1,X2$ be independent random variables with $P[Xi = 0] = P[Xi = 1] = \frac{1}{2}$ for $i = 1, 2$. What is the random variable set $A1 = \{X1 = X2\}$ mean? I came across this representation in the book but not sure what this means.
 A: Although this question may look very simple, a full understanding of it is not. For general purposes, I give the following elaborated answer.
In general, if $X_1$ and $X_2$ are any two random variables on a common probability space $(\Omega,\mathcal{F},{\rm P})$, then $\lbrace X_1 = X_2 \rbrace$ stands as a shorthand for the event $\lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace$. For your specific example, we can define $(\Omega,\mathcal{F},{\rm P})$ as follows:
$\Omega = \lbrace (0,0),(0,1),(1,0),(1,1) \rbrace$, $\mathcal{F}=2^\Omega$ (the power set of $\Omega$), and, for any $\omega = (i,j) \in \Omega$, ${\rm P}(\lbrace \omega \rbrace) = 1/4$. Note that, by additivity, this determines (the probability measure) ${\rm P}$ on $\mathcal{F}$; for example,
$$
{\rm P}(\lbrace (0,0),(1,0),(1,1)\rbrace) = {\rm P}(\lbrace (0,0)\rbrace) + {\rm P}(\lbrace (1,0)\rbrace) + {\rm P}(\lbrace (1,1)\rbrace) = 3/4.
$$
Now, we can define the random variables $X_1$ and $X_2$ as follows. For any $\omega = (i,j) \in \Omega$, $X_1 {(\omega)} = i$ and $X_2 {(\omega)} = j$. Thus, 
$$
{\rm P}(X_1 = 0) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = 0 \rbrace) = {\rm P}(\lbrace (0,0),(0,1) \rbrace) = 1/2
$$
and
$$
{\rm P}(X_1 = 1) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = 1 \rbrace) = {\rm P}(\lbrace (1,0),(1,1) \rbrace) = 1/2,
$$
and analogously for $X_2$. As for the event in question,
$$
\lbrace X_1 = X_2 \rbrace : = \lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace = \lbrace (0,0),(1,1) \rbrace,
$$
and so
$$
{\rm P}(X_1 = X_2) := {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = X_2 {(\omega)} \rbrace) = {\rm P}(\lbrace (0,0),(1,1) \rbrace) = 1/2.
$$
Finally, let's see that $X_1$ and $X_2$ are indeed independent (this should be clear from our construction). This amounts to showing that, for any $i,j \in \lbrace 0,1 \rbrace$, 
$$
{\rm P}(X_1 = i, X_2 = j) = {\rm P}(X_1 = i) {\rm P}(X_2 = j), 
$$
that is,
$$
{\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} = i, X_2 {(\omega)} = j \rbrace) = {\rm P}(\lbrace \omega \in \Omega : X_1 {(\omega)} =i  \rbrace ){\rm P}(\lbrace \omega \in \Omega : X_2 {(\omega)} =j  \rbrace ).
$$
Indeed, both sides are equal to $1/4$.
A: To add to Shai Covo's answer, it's worth noting that the convention of writing something like $\{X_1 = X_2\}$ for $\{\omega \in \Omega : X_1(\omega) = X_2(\omega)\}$, essentially "suppressing the $\omega$s", is very common in probability, and you will see it a lot as you continue your studies.  Essentially, any "statement" written inside braces should be interpreted as the set of all $\omega$ for which the statement (which is generally something in terms of random variables, which are really functions of $\omega$) is true.  Other common examples are things like $\{X_n \to X\}$, which is the set of all $\omega$ for which $X_n(\omega)$ converges to $X(\omega)$.  At first you may find it helpful to "fill in the $\omega$s" when reading.
However, for intuition, it is often easy to understand such expressions.  If you are doing an experiment where you flip two coins, with heads and tables labeled 0 and 1, $\{X_1 = X_2\}$ is just the event that the two coins come up the same.
A: It's the event that $X_1 = X_2$.  So $X_1 = 0, X_2 = 0$ or $X_1 = 1, X_2 = 1$. 
