Let $\mathcal A$ be an algebra on a set $X$, $\mu_0$ a premeasure on $\mathcal A$,
and $\mathcal M$ the $\sigma$-algebra generated by $\mathcal A$.
Let $\mu^*$ be the Caratheodory outer measure induced by $\mu_0$, $\mu$ the restriction of $\mu^*$ on $\mathcal M$.
Let $\mathcal C$ be the set of all $\mu^*$-measurable sets, $\bar{\mu}$ the restriction of $\mu^*$ on $\mathcal C$.
Suppose $\mu_0$ is $\sigma$-finite, i.e. there exists a sequence $E_n, n = 1, 2, \cdots$ of members of $\mathcal A$ such that $X = \cup_{n=1}^{\infty} E_n$, $\mu_0(E_n) \lt \infty$ for all $n$. I claim that $(X, \mathcal C, \bar{\mu})$ is the completion of $(X, \mathcal M, \mu)$.
Let $\bar{\mathcal M}$ be the completion of $\mathcal M$ with respect to $\mu$.
Since $\mathcal M\subset \mathcal C$, $\bar{\mathcal M}\subset \mathcal C$.
Hence it suffices to prove that $\mathcal C \subset \bar{\mathcal M}$.
Since $\mu_0$ is $\sigma$-finite, it suffices to prove that if $E\in \mathcal C$ and $\bar{\mu}(E) \lt \infty$, then $E\in \bar{\mathcal M}$.
By the definition of $\bar{\mu}$, for each integer $n \ge 1$, there is a sequence $A_j, j= 1, 2, \cdots$ such that $\sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$, where $A_j \in \mathcal{A}$, $E \subset \bigcup_{j=1}^\infty A_j.$ Let $F_n = \bigcup_{j=1}^{\infty} A_j$. Then $E \subset F_n$ and $\mu(F_n) \le \sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$.
Let $F = \cap_{n=1}^{\infty} F_n$. Then $F\in \mathcal M$, $E \subset F$, and $\bar{\mu}(E) \le \mu(F) \le \mu(F_n) \lt \bar{\mu}(E) + 1/n$ for all $n\ge 1$. Hence $\bar{\mu}(E) = \mu(F)$.
Similarly there exists $G\in \mathcal M$ such that $F - E \subset G$ and $\mu(G) = \bar{\mu}(F - E)$ = 0.
Then $E = (F - G) \cup (E\cap G)$, $F - G \in \mathcal M$, and $E\cap G$ is a subset of the $\mu$-null set $G$. Hence $E \in \bar{\mathcal M}$. This completes the proof.
Now consider the family $\mathcal A$ of finite disjoint unions of intervals of the form $[a, b)$ or $(-\infty, c)$, where $-\infty \lt a\lt b\le \infty$ and $c\in \mathbb R$.
It is elementary and well known that $\mathcal A$ is an algebra on $\mathbb R$ and there is a unique premeasure $\mu_0$ on $\mathcal A$ such that $\mu_0([a, b)) = b - a$ whenever $a$ and $b$ are finite.
Then $\mathcal M$ and $\mathcal C$ defined above are the families of Borel sets and Lebesgue measurable sets in $\mathbb R$ respectively and you get the picture.
The set of Borel sets is the smallest collection of sets that contains the open sets and is closed under countable unions and intersections and complements. The set of Lebesgue measurable sets is the smallest collection of sets that contains the open sets and that is closed under countable unions and intersections and complements and which is such that for any set of measure $0$, any subset of that set is measurable. Because Lebesgue sets have this property of measure $0$ sets, we say that the $\sigma$-algebra is complete.
So, Borel sets are Lebesgue sets, but not vice versa. Borel sets don't need a measure, they just need a topology, but Lebesgue sets need a measure to complete the $\sigma$-algebra.
Using the definitions you presented, you should be able to answer your own question now.
Best Answer
Good question; I didn't know the answer until just now, never having thought about it.
A: Because if we did there would be fewer measurable functions, and in particular a continuous function $f:\Bbb R\to\Bbb R$ would not necessarily be measurable:
Say $K$ is a "fat Cantor set", that is a subset of $\Bbb R$ homeomorphic to the middle-thirds Cantor set $C$ but such that $m(K)>0$. There is a continuous bijection $f:\Bbb R\to\Bbb R$ such that $f(K)=C$. Now $K$ contains a non-measurable set $E$. Say $F=f(E)$. Then $F$ is Lebesgue measurable, being a subset of a null set, but its inverse image $E$ is not Lebesgue measurable. So $f$ is continuous but not Lebesgue-to-Lebesgue measurable.
Not what we want; if continuous functions are not measurable none of the things we want to use this stuff for work anymore.
(Also it's fun to ask students to resolve the "paradox" that according to the definition of a measurable function between measurable spaces, no topology, the composition of two measurable functions is obviously measurable, yet in this context the composition of two measurable functions need not be measurable.)