Consider: $$S \subset S^*$$
I'd like to prove that $S$ is a Borel set up to a set of measure zero. I know $S$ is a set of full measure and $S^*$ is a Borel set. The logic in the video (linked below) proceeds as follows:
$S$ is full measure, thus $S^*$ is full measure, which implies $S$ is Borel up to measure zero set.
I'm confused because:
In order to know $S$ is a set of full measure, it needs to be in the sigma-algebra associated with Leb measure. So why did he need the bigger set $S^*$ which is Borel? If you are a set of full measure with respect to Leb measure, aren't you automatically a Borel set?
In his explanation, he says the compliment of $S^*$ is a set of zero measure so the compliment of $S$ is a set of zero measure. This is only true if $S^*-S$ (the compliment of $S$ in $S^*$) has zero measure, right?
If $S$ is full measure and $S \subseteq S^*$, then necessary $S^*$ is also set of full measure: if $\mu(\overline{S^*}) \leqslant \mu(\overline{S}) = 0$. And also $S$ is full measure in $S^*$: $\mu(S^* \setminus S) \leqslant \mu(\overline{S}) = 0$. So $S$ is a Borel set $S^*$ up to set of zero measure $S^* \setminus S$.
It's not clear why we need $S^*$ here - we can just take the whole space $X$ (that is Borel) instead.
And no, not all full measure sets are Borel. Take any zero set $T$ that is Lebesgue measurable but not Borel - then $\overline{T}$ has full measure, but is not Borel.