Dudley, in " Real Analysis and Probability", defines and proves the recursion principle as it follows:
The passage tagged in yellow is an application of the induction principle, which has been aforedefined in the text as:
I didn't understand how the induction principle implies the highlighted passage. Could somebody please explain it to me?
$J(x)$ is well-ordered, we want to show that the subset $$ Y := \left\{u \in J(x) \mid h(u) = f(u) \right\} $$ in the whole of $J(x)$ (that is the highlighted statement). By (1.3.1) it suffices to show that $Y$ is inductive: We have $b \in Y$ as $$ f(b) = g(\emptyset) = h(b) $$ Now suppose $u \in J(x)$ such that $I(u) \subseteq Y$ is given. Then, by the properties of $h$ and $f$ $$ f(u) = g\bigl(f|_{I(u)}\bigr), \qquad h(u) =g\bigl(h|_{I(u)}\bigr) $$ As $I(u) \subseteq Y$, we have $f|_{I(u)} = h_{I(u)}$, hence $$ f(u) = g\bigl(f|_{I(u)}\bigr) = g\bigl(h|_{I(u)}\bigr) = h(u) $$ that is $u \in Y$. Hence $Y$ is inductive and therefore, by (1.3.1), $Y = J(x)$.
