Lower bound of disjointness by discrepancy? I need to show that $Disc_\mu(Disj) \geq \frac{1}{2n+1}$ for any distribution $\mu: \{0,1\}^n \times \{0,1\}^n \to [0,1]$. 
Disjointness is defined as 

$Disj(X,Y)=\left\{ \begin{array}[ll]+1 &amp; \text{if $X \cap Y = \emptyset$} \\ 0 &amp; \text{otherwise}\end{array}\right.$

The discrepancy for a function is defined as  $Disc_\mu(f) = \max_R \{Disc_\mu(R,f)\}$ over all rectangles $R$ and further

$$Disc_\mu(R,f) = |\Pr_\mu [(x,y) \in R \wedge f(x,y)=1] - \Pr_\mu[(x,y) \in R \wedge f(x,y) =0] |.$$

I don't understand where the $\frac{1}{2n+1}$ comes from and how to define a proper rectangle $R$.

The problem is given as an exercise without a solution in the book Communication Complexity by E. Kushilevitz and N. Nisan on page 40.
You asked two questions.  I'll answer the second.  A rectangle is a set $R$ of the form

$$R = \{(x_1,\dots,x_n) : \ell_1 \le x_1 \le u_1, \dots, \ell_n \le x_n \le u_n\}$$

for some $\ell_1,\dots,\ell_n,u_1,\dots,u_n$.