What is the difference between a measurable set and a $\mu^*$-measurable set?

In measure theory class, we learned that a set $B$ is $\mu^*$-measurable. If $\forall T\subseteq \Omega$, we have $$\mu^*(T)=\mu^*(T\cap B)+\mu^*(T\cap B^c)$$

Then to prove that the Lebesgue measure space is complete (Proof 1 and Proof 2) we did the following. Let $\mu_l$ the lebesgue measure and $\mu_l(A)=0$ and $B\subseteq A$ we used the if $\mu^*_l$ is the outer Lebesgue measure then $$\mu_l^*(T)=\mu_l^*(T\cap B)+\mu_l^*(T\cap B^c)$$

and hence $B$ is Lebesgue measurable.

This where my confusion stared.

Does "Lebesgue measurable" mean "$\mu_l$-measurable" or "$\mu^*_l$-measurable"?

Since $$\mu_l^*(T)=\mu_l^*(T\cap B)+\mu_l^*(T\cap B^c)$$ only showed that the $B$ is measurable with respect to the outer lebesgue measure. Why does this mean the $B$ is Lebesgue measurble?

Is every outer-measurable set measurable?

It is by definition. If $B$ satisfies the Carathéodory criterion $\mu^*(T)=\mu^*(T\cap B)+\mu^*(T\cap B^c)$ for every $T\subset\mathbb{R}$, then $B$ is Lebesgue measurable and we may define $\mu(B)=\mu^*(B)$. Yes, you can call $B$ $\mu^*$-measurable, but this just refers to the outer measure used in the Carathéodory criterion (in case, for example, you are working with more than one outer measure). In general, you just say $B$ is (Lebesgue) measurable.

Is every outer-measurable set measurable?

Again, if by this you mean $B$ satisfies the Carathéodory criterion, then yes. If by this question you mean a set $B\subset\mathbb{R}$ such that $\mu^*(B)$ makes sense, then no. Remember, outer measure is defined for all sets in $\mathbb{R}$, and there are certainly sets for which the Carathéodory criterion does not hold.