Uniqueness of Lebesgue measure

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I am studying measure theory using the lectures by Claudio Landim on Youtube.

In lecture 4 he proofs the Caratheodory extension theorem and in particular points out that it is possible to extend a $\sigma$-additive set function defined on an algebra $a$ to a $\sigma$-additive set function defined on the sigma algebra of measurable sets $M$ which satisfy the Caratheodory measurability criterion, i.e. $A \in M$ if $$\pi^*(E)=\pi^*(E \cap A)+\pi^*(E \cap A^c)$$ where $\pi^*$ denotes the Lebesgue outer measure.

$M$ is usually called the Lebesgue $\sigma$-algebra and the restriction of the Lebesgue outer measure to the set $M$ is simply referred to as the Lebesgue measure (see for instance https://en.wikipedia.org/wiki/Lebesgue_measure#Definition).

However, he also states that we can only prove the uniqueness of the Lebesgue measure on a possibly smaller $\sigma$-algebra $\varphi(a)$ (the $\sigma$-algebra generated by the algebra) under certain conditions I will omit for brevity. Note that $\varphi(a) \subseteq M$ since $M$ contains $a$.

So to sum up we should always consider the Lebesgue measure on $\varphi(a)$ since otherwise it is not clear what we mean by $\pi^*(A)$ for a set $A$.

In lecture 5 he continues by constructing the Lebesgue measure on $\mathbb{R}$ by showing that we can apply the Caratheodory criterion to a $\sigma$-additive set function defined on a suitable algebra.

Now the thing that confuses me is that most theorems and definitions using the Lebesgue measure consider the Lebesgue $\sigma$-algebra and not $\varphi(a)$.

For example, Lebesgue measurable functions are defined as functions such that the inverse image $f^{-1}(A) \in M$ $\forall A \in B$ where $B$ is the Borel $\sigma$-algebra (https://en.wikipedia.org/wiki/Measurable_function#Notable_classes_of_measurable_functions).

Why do we always consider the Lebesgue measure on $M$ when it is not always unqiue? Or is the Lebesgue measure on $\mathbb{R}$ a special case? This has been confusing me for quite a while now, so any help is greatly appreciated.