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Inspecting a proof of the $\pi$–$\lambda$ theorem, it seems to me that one instance of transfinite recursion/induction is the following:
As part of the proof we need to show that a $\pi$–$\lambda$ system $F\subseteq 2^\Omega$ is a $\sigma$-algebra. We want to show that if $F$ is closed under countable disjoint unions and finite intersections then it is closed under arbitrary countable unions. Suppose $X_i\in F$ where $i\in\omega$, we want to show that $\bigcup_{i\in\omega}{X_i}\in F$ by showing that there are disjoint sets $Y_i\in F$, $i\in\omega$ such that $\bigcup Y_i=\bigcup X_i$.
From the set-theoretic point of view, this means we want to assert the existence of a function $G: \omega\rightarrow 2^\Omega$ s.t. $G(i)=Y_i$, which is precisely what (a weak instance of) the theorem of transfinite recursion gives you. This theorem tells you that having constructed $G|_n$, if you define $G(n+1)$ in terms of $G|_n$, then you can proceed to get the whole $G$ defined on $\omega$. In this case at these successor steps we just remove the intersections, which we can do by closure under finite intersection.
Note that in this case, we don’t need the full transfinite recursion, which says if you also specify what the construction does at limit stages then you can get functions defined on $\alpha$ where $\alpha$ is any ordinal.
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