4. Define a tetrahedron with equal opposite edges as an isosceles tetrahedron. Let the circumradius of the isosceles tetrahedron DBMN be $R$, and the circumradius of $\triangle B M N$ be $r$. Given
$$
D B=M N=a, D M=B N=b, D N=B M=c \text {. }
$$

then the range of $\frac{r}{R}$ is $\qquad$
4. $\left[\frac{2 \sqrt{2}}{3}, 1\right)$.

By definition, a tetrahedron with equal opposite edges can be completed into a rectangular prism $A B C D-M K N L$, as shown in Figure 2. Then the tetrahedron $D B M N$ is a tetrahedron with equal opposite edges, and its circumscribed sphere diameter is the diagonal $D K$ of the rectangular prism.
Let the height, length, and width of the rectangular prism be $A M=x, A B=y, A D=z$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
x^{2}+y^{2}=c^{2}, \\
y^{2}+z^{2}=a^{2}, \\
z^{2}+x^{2}=b^{2}
\end{array}\right. \\
\Rightarrow R=\frac{1}{2} \sqrt{x^{2}+y^{2}+z^{2}} .
\end{array}
$$

For an isosceles tetrahedron, all four lateral faces are the same. Using Heron's formula, we get
$$
\begin{array}{l}
S_{\triangle B M N}=\frac{1}{2} \sqrt{y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}} \\
\Rightarrow r=\frac{\sqrt{\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)\left(x^{2}+y^{2}\right)}}{2 \sqrt{y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}}} \\
\Rightarrow \frac{r}{R}=\frac{\sqrt{\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)\left(x^{2}+y^{2}\right)}}{\sqrt{\left(x^{2}+y^{2}+z^{2}\right)\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right)}} .
\end{array}
$$

By the identity
$$
\begin{array}{l}
\left(x^{2}+y^{2}+z^{2}\right)\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right) \\
=\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)\left(x^{2}+y^{2}\right)+x^{2} y^{2} z^{2},
\end{array}
$$

and the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{8}{9} \leqslant \frac{\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right)\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}+z^{2}\right)\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right)}<1 \\
\Rightarrow \frac{2 \sqrt{2}}{3} \leqslant \frac{r}{R}<1
\end{array}
$$