Context: This is from Chapter 3 of Christodoulou and Klainerman's stability of Minkowski space. We have a family of metrics $m_u$ on $S^2$ for each $u \in (u_0, \infty)$ (for some fixed $u_0 < 0$). We also define, separately, $m_\infty$ to be the standard unit round metric on $S^2$. The goal is to prove, in an appropriate sense, that $(S^2, m_u)$ converges to $(S^2, m_\infty)$ (under some additional assumptions, the most geometrically illuminating of which is that $K(u) \to 1$ as $u \to \infty$, where $K(u)$ is the Gaussian curvature of $m_u$.
Let $(e_A)_{A = 1, 2}$ be an $m_\infty$-ONB such that the matrix $m_u(e_A, e_B)$ is diagonal, with smallest eigenvalue $\lambda(u)$ and largest eigenvalue $\Lambda(u)$. The goal is to show that $\lambda(u), \Lambda(u) \to 1$ as $u \to \infty$.
Notation: We let $r(u)$ be the "radius" of $(S^2, m_u)$, defined by the formula $$ \text{Area}(S^2, m_u) = 4\pi r(u)^2. $$ Define $\mu(u) = \sqrt{\det_{m_\infty} m_u(e_A, e_B)} = \sqrt{\lambda(u)\Lambda(u)}$.
Assumptions: 1) The following integral is finite, and uniformly bounded for all values of $u$: $$ \int_u^\infty r(u')^{-1} \kappa(u')\, du'. $$ Here $\kappa$ is some nonnegative scalar function.
One can derive the following ODE for $\mu$: $$ \partial_u \mu(u) = r^{-1}\kappa \mu. $$ C-K then say that $$ \mu(u) = \exp(-\int_u^\infty r(u')^{-1}\kappa(u')\, du'). $$ My question: There should be a constant out front corresponding to the "initial" value $\lim_{u \to \infty}\mu(u)$. Hence it appears they are saying it is 1. I don't see how this is possible, as this seems to be what we are trying to prove?