4. Given
$$
\begin{array}{l}
M=\left\{(x, y) \mid y \geqslant x^{2}\right\}, \\
N=\left\{(x, y) \mid x^{2}+(y-a)^{2} \leqslant 1\right\} .
\end{array}
$$

Then, the necessary and sufficient condition for $M \cap N=N$ to hold is:
(A) $a \geqslant 1 \frac{1}{4}$;
(B) $a=1 \frac{1}{4}$;
(C) $a \geqslant 1$;
(D) $0<a<1$.
4. (A)

As shown in the figure, $M$ represents the region inside the parabola $y=x^{2}$ including the boundary, and $N$ represents the circular region with center at $(0, a)$ and radius 1 (including the circumference). Its equation is
$$
x^{2}+(y-a)^{2}=1 \text {. }
$$

The necessary and sufficient condition for $M \cap N=N$ to hold is $M \subseteq N$. That is, the circular region $N$ is inside the parabola $y=x^{2}$, and can have common points with the boundary curve $y=x^{2}$.

Let $(x, y)$ be any point on the parabola $y=x^{2}$. From the distance relationship between it and the center $(0, a)$, the required necessary and sufficient condition is
$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + ( y - a ) ^ { 2 } \geqslant 1 } \\
{ y = x ^ { 2 } } \\
{ a \geqslant 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y^{2}-(2 a-1) y+a^{2}-1 \geqslant 0 \\
y=x^{2} \\
\alpha \geqslant 1 .
\end{array}\right.\right.
$$

For any $y \in R$, the inequality $y^{2}+(1-2 a) y+(a-1) \geqslant 0$ always holds.
$$
\therefore \quad \Delta=(2 a-1)^{2}-4\left(a^{2}-1\right) \leqslant 0
$$

Solving this, we get
$$
a \geqslant \frac{5}{4} \text {. }
$$
$\therefore$ The necessary and sufficient condition for $M \cap N=N$ is $a \geqslant 1 \frac{1}{4}$. Choose (A).
In fact, the correct answer (A) can also be determined by intuition or elimination. From the figure, $N$ is the circle with center $(0, a)$ and radius 1, including the circumference. When $a$ is a sufficiently large positive number, it is clear that $N \subset M$. Thus, options (B) and (D) are eliminated.

When $a=1$, it is easy to find that the parabola $y=x^{2}$ and the circle $x^{2}+(y-1)^{2}=1$ intersect at three points $(0,0),(1,1),(-1,1)$,
Therefore, $N \not \subset M$, eliminating (C).
Since in each sub-question, among the four options (A), (B), (C), (D), "there is one and only one correct answer".
Thus, the correct answer must be (A).