4. Let in the $x O y$ plane, the area enclosed by $0<y \leqslant x^{2}, 0 \leqslant x \leqslant 1$ be $\frac{1}{3}$. Then the intersection of the sets
$$
\begin{array}{l}
M=\{(x, y)|| y|-| x \mid \leqslant 1\}, \\
N=\left\{(x, y)|| y \mid \geqslant x^{2}+1\right\}
\end{array}
$$

representing the area of $M \cap N$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{2}{3}$
(C) 1
(D) $\frac{4}{3}$
4.B.
$M \cap N$ is symmetric about both the $x$-axis and the $y$-axis in the $x O y$ plane. Therefore, the area of $M \cap N$ can be calculated by multiplying the area in the first quadrant by 4. For this purpose, we only need to consider the area in the first quadrant. According to the problem, the area of $M \cap N$ in the first quadrant is $A=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$. Thus, the area of $M \cap N$ is $\frac{2}{3}$.