Working with measures over $\sigma$-algebras is a recurring theme over the whole book. The very first problem one encounters doing it, is that $\sigma$-algebras are unmanageably huge! so when for example you want to show that two measures $\nu_1$ and $\nu_2$ are equal on $\mathcal{F}$, you want to avoid testing their equality on every single member of the $\mathcal{F}$. Keeping that in mind, would make reasonable all the difficulties of starting from algebras then extending to semi-algebras and then extending to $\sigma$-algebras. To get a sense of how surprisingly large a $\sigma$-algebra is, look at \textbf{Remark 3.10} and the discussion after it in
The answer in \citet{mathstackdynkin} is also related as it explains how the $\pi-\lambda$ theorem helps us avoid working directly with $\sigma$-algebras. Plus, it clarifies how the seemingly ad-hoc definition of $\lambda$-system makes sense when we want to investigate agreement of two measures on individual elements of a $\sigma$-algebra.
Theorem A.1.4. (\(\pi-\lambda\) Theorem) One of the most used theorems in the book. Recall that by the definition, any $\sigma$-algebra is automatically a $\lambda$-system. Keeping that in mind, we want to look if the generated $\sigma$-algebra from a generator inside a $\lambda$-system will stay in that $\lambda$-system or not! (Spoiler: the answuer is YES!)
This theorem says: consider you have a collection of sets closed under intersection as the generator of a $\sigma$-algebra. If the generator is contained in a $\lambda$-system, then the generated $\sigma$-algebra will remain in the $\lambda$-system.
But how this result helps us on our initial motivation to avoid working directly with every single member of the $\sigma$-algebra? Actually when we look at its use-case in Theorem A.1.5 it will be more clear. There, we show the uniqueness of a measure for a defined $\lambda$-system, then by knowing that the $\lambda$-system covers the generator, using the result of $\pi-\lambda$ theorem we conclude that the agreement of measures is also true for the generated $\sigma$-algebra, as the generated $\sigma$-algebra is contained in the $\lambda$-system.
Theorem A.1.5 This theorem shows uniqueness of measure defined on a $\pi$-system will result in its uniqueness also on the $\sigma$-algebra generated by that $\pi$-systems. Now you might wonder that why we are trying to show this, while we are interested in uniqueness of measures on the $\sigma$-algebra generated by an algebra, as mentioned in the Theorem A.1.3. Carath'eodory’s Extension Theorem. The answer is that the any algebra is automatically a \(\pi\)-system, by definition, therefore by this theorem we are proving the desired result for a larger collection of sets (if it’s true for any $\pi$-system then it’s true for any algebra).
First we focus on a collection of sets in the generated $\sigma$-algebra on whom the measures agree and also covers the whole initial $\pi$-system/generator. The idea then is, if we can show that this collection of sets in the generated $\sigma$-algebra is a $\lambda$-system, then using the $\pi-\lambda$ theorem we would say that the $\sigma$-algebra is included in the $\lambda$-system. On the other hand, as we defined the $\lambda$-system to have all the sets that the two measures agree on them, therefore for any set in the generated $\sigma$-algebra the measures agree as well.