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

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
$$
\left\{\begin{array}{l}
x^{2}+(y-a)^{2} \geqslant 1 \\
y = x^{2} \\
a \geqslant 1
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y^{2}-(2 a-1) y+a^{2}-1 \geqslant 0 \\
y = x^{2} \\
a \geqslant 1 .
\end{array}\right.\right.
$$

For any $y \in \mathbb{R}$, 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 must be 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 $x^{2}+(y-1)^{2}=1$ have three intersection points $(0,0),(1,1),(-1,1)$,
Therefore, $N \not \subset M$, eliminating (C).
Since among the four options (A), (B), (C), (D) for each sub-question, "there is one and only one correct answer".
The correct answer must be (A).