enter image description here

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?

Thank you so much for your valuable time for reading.


1 Answer 1


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.

  • $\begingroup$ Oh! Sorry I messed up Vitali set with Cantor set. How silly am I! Thanks for pointing that out. $\endgroup$ Jun 11, 2020 at 16:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .