4. Given a unit cube $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, there is a point $M$ on the side $A A^{\prime} B^{\prime} B$ such that the distances from $M$ to the lines $A B$ and $B^{\prime} C^{\prime}$ are equal. Then the minimum distance from $M$ to point $C^{\prime}$ is $\qquad$
4. $\frac{\sqrt{5}}{2}$.

Taking the midpoint $O$ of $B B^{\prime}$ as the origin, the direction of $B A$ as the $x$-axis, and the direction of $B B^{\prime}$ as the $y$-axis, establish a Cartesian coordinate system.
Then, according to the problem, the trajectory of point $M$ is
$$
y=\frac{1}{2} x^{2}(0 \leqslant x \leqslant 1),
$$

The minimum distance from a point on this trajectory to point $B^{\prime}$ is $\frac{1}{2}$.
Therefore, the minimum distance from the moving point $M$ to point $C^{\prime}$ is $\frac{\sqrt{5}}{2}$.