4. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, $A_{1} B \perp$ $B_{1} C$. Then the angle formed by $A_{1} B$ and the plane $B B_{1} C_{1} C$ is equal to ( ).
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $\arctan \sqrt{2}$
(D) $90^{\circ}$
4.A.

As shown in Figure 1, let $M$ be the midpoint of $AB$, and connect $CM$ and $B_{1}M$. Clearly, $CM \perp$ plane $AA_{1}B_{1}B$.

By the theorem of three perpendiculars, $B_{1}M \perp A_{1}B$. Therefore, 
$\triangle B_{1}BM \sim \triangle A_{1}B_{1}B$.
Thus, $A_{1}B_{1}: BB_{1} = \sqrt{2}: 1$.
Let $N$ be the midpoint of $B_{1}C_{1}$, then $A_{1}N \perp$ plane $BB_{1}C_{1}C$. Therefore, $\angle A_{1}BN$ is the angle between $A_{1}B$ and plane $BB_{1}C_{1}C$.
Let $BB_{1} = a$, then $A_{1}B_{1} = \sqrt{2}a$.
Thus, $A_{1}B = \sqrt{3}a$, $A_{1}N = \frac{\sqrt{6}a}{2}$.
Therefore, $\sin \angle A_{1}BN = \frac{A_{1}N}{A_{1}B} = \frac{\sqrt{2}}{2}$.