# Isn't the Lebesgue measure space complete? Let $$X$$ be a non-empty set. Let $$\mathcal A$$ be an algebra of subsets of $$X$$ and $$\mathcal S (\mathcal A)$$ be the $$\sigma$$-algebra of subsets of $$X$$ generated by $$\mathcal A.$$ Let $$\mu : \mathcal A \longrightarrow [0,+\infty]$$ be a measure on $$\mathcal A.$$ Let $$\mu^*$$ be the induced outer measure. Let $$\mathcal S^*$$ be the $$\sigma$$-algebra of $$\mu^*$$-measurable subsets of $$X.$$ Then what I know is that $$\mathcal A \subseteq \mathcal S^*$$ and hence $$\mathcal S (\mathcal A) \subseteq \mathcal S^*.$$ Then the measure space $$(X,\mathcal S^*, \mu^*)$$ is complete since $$\mathcal N \subseteq \mathcal S^*,$$ where $$\mathcal N : = \{E \subseteq X\ |\ \mu^*(E) = 0 \}.$$ The measure space $$(X,\mathcal S^*,\mu^*)$$ is called the completion of the measure space $$(X,\mathcal S (\mathcal A),\mu^*).$$

For the Lebesgue measure space we have $$X = \Bbb R,$$ $$\mathcal S^* = \mathcal L_ {\Bbb R},$$ the $$\sigma$$-algebra of Lebesgue measurable sets, $$\mathcal S(\mathcal A) = \mathcal B_{\Bbb R},$$ the $$\sigma$$-algebra of Borel sets and $$\mu^* = \lambda^*,$$ the outer Lebesgue measure induced by the length function. Hence in this case we can say that the Lebesgue measure space $$(\Bbb R, \mathcal L_{\Bbb R}, \lambda^*)$$ is complete and it is the completion of $$(\Bbb R, \mathcal B_{\Bbb R},\lambda^*).$$ Let $$\mathcal N : = \{E \subseteq X\ |\ \lambda^*(E)=0 \}.$$ Now since the Lebesgue measure space is complete, $$\mathcal N \subseteq \mathcal L_{\Bbb R}.$$ That means all the subsets of $$\Bbb R$$ which have outer Lebesgue measure $$0$$ are Lebesgue measurable. But how can it be true in reality? I know the existence of non-Lebesgue measurable sets (i.e. Vitali set) having outer Lebesgue measure $$0.$$ I don't understand where did I mess up! Can anybody please help me in clearing my confusion?

I know the existence of a non-Lebesgue measurable sets (i.e. Vitali set) having outer Lebesgue measure $$0$$
That's not true. Vitali sets have positive outer measure, and indeed that's how we prove they're not measurable (if they were, by the countable additivity of Lebesgue measure the interval $$[0,1]$$ would have to have infinite measure). It is indeed the case that all outer-measure-zero sets are measurable.