In most measure theory text books one derives the Lebesgue anh Borel-Lebesgue measure from Caratheod

In most measure theory text books one derives the Lebesgue anh Borel-Lebesgue measure from Caratheodory's extension to outer measures by first proving that the set of ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ measurable sets is a sigma algebra (the Lebesgue sigma algebra) and that ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ restricted to that sigma algebra is a measure, the Lebesgue measure. This measure space is even complete. However, then still one continues to show that ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ restricted to the sigmal algebra generated by the ring (on which the pre-measure was defined that was used to obtain ${\mathrm{\lambda <!--\; \lambda \; -->}}^{\ast <!--\; \ast \; -->}$ via Caratheodorys extension) is also a measure, the Borel-Lebesgue measure, and that the generated sigma algebra is the Borel sigma algebra.
So my question is, why is the Borel sigma-algebra "better" than the Lebesgue sigma algebra, because most of the time text books continue to work only on the Borel sigma algebra, even though the Lebesgue sigma algebra is its completetion and has some other favorable properties? I.e., I am just missing an argument in all the lecture notes and text books why we continue to work on the Borel sigma-algebra after having shown that the Lebesgue sigma algebra is larger (and after all we do all the extension from a pre-measure to an outer measure and then restricting to a measure because we want to get a larger set than just the ring on which we initially defined the pre-measure).