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?
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