4. As shown in the figure, in trapezoid $A B C D$, $A B / / C D$, and $A B=2 C D, M 、 N$ are the midpoints of diagonals $A C 、 B D$. Let the perimeter of trapezoid $A B C D$ be $l_{1}$, and the perimeter of quadrilateral CDMN be $l_{2}$. Then $l_{1}$ and $l_{2}$ satisfy $(\quad)$.
(A) $l_{1}=2 l_{2}$
(B) $l_{1}=3 l_{2}$
(C) $l_{2}<l_{1}<2 l_{2}$
(D) $2 l_{2}<l_{1}<3 l_{2}$
4. (A).

As shown in the figure, let the line containing $MN$ intersect $AD$ and $BC$ at $P$ and $Q$, respectively. Then $P$ and $Q$ are the midpoints of $AD$ and $BC$, respectively. Therefore,
$$
\begin{aligned}
P M & =N Q=\frac{1}{2} C D . \\
\therefore P Q & =P M+M N+N Q=C D+M N \\
& =\frac{1}{2}(A B+C D) .
\end{aligned}
$$

Extend $DM$ to intersect $AB$ at $R$, then $P M=\frac{1}{2} A R=\frac{1}{2} C D$.
$$
\because A B=2 C D, \therefore A R=C D=R B \text {. }
$$

Therefore, $BCDR$ is a parallelogram.
Thus, $D M=\frac{1}{2} D R=\frac{1}{2} B C$.
Similarly, $C N=\frac{1}{2} A D$.
$$
\therefore D M+C N=\frac{1}{2}(B C+A D) \text {. }
$$

From (1) and (2), we get $l_{2}=\frac{1}{2} l_{1}$, i.e., $l_{1}=2 l_{2}$.