Vitali coverings on the real line

Jordan Bell
March 10, 2016

For x and r>0 write


Let λ be Lebesgue measure on the Borel σ-algebra of and let λ* be Lebesgue outer measure on .

A Vitali covering of a set E is a collection 𝒱 of closed intervals such that for ϵ>0 and for xE there is some I𝒱 with xI and 0<λ(I)<ϵ.

The following is the Vitali covering theorem.11 1 Klaus Bichteler, Integration – A Functional Approach, p. 161, Lemma 10.5; John J. Benedetto and Wojciech Czaja, Integration and Modern Analysis, p. 179, Theorem 4.3.1; Russell A. Gordon, The Integrals of Lebesgue, Denjoy, Perron, and Henstock, p. 52, Lemma 4.6.

Theorem 1 (Vitali covering theorem).

Let U be an open set in R with λ(U)<, let EU, and let V be a Vitali covering of E each interval of which is contained in U. Then for any ϵ>0, there are disjoint I1,,InV such that


Suppose that I1,,In𝒱 are pairwise disjoint. If Ej=1nIj then I1,,In satisfy the claim, and otherwise, let


and there exists some xEUn. As xUn and Un is open, there is some η>0 such that B(x,η)Un and then as 𝒱 is a Vitali covering of E there is some I𝒱 with xIB(x,η)Un. Thus δn>0 for


and there is some In+1𝒱 with In+1Un and λ(In+1)>δn2.

For j1 write Ij=[xj-rj,xj+rj] and let Jj=[xj-5rj,xj+5rj], namely Jj is concentric with Ij and λ(Jj)=5λ(Ij). Then, as the intervals I1,I2, are pairwise disjoint Borel sets each contained in U,


and it follows from j=1λ(Jj)< that j=Mλ(Jj)0 as M, which with


yields λ(j=MJj)0 as M.

Let M1. If xEj=1Ij then xEj=1MIj and so xUM, and as UM is open there is some η>0 with B(x,η)UM. But xE and 𝒱 is a Vitali covering of E, so there is some I𝒱 with xI and IB(x,η)UM. Now, λ(Ij+1)>δj2 and j=1λ(Ij)< together imply δn0 as n, so there is some n for which δn<λ(I). By the definition of δn as a supremum, this means that IUn and so it makes sense to define N to be a minimal positive integer such that IUN. M<N: if MN then IUMUN, contradicting IUN. (We shall merely use that MN.) The fact that IUN and IUN-1 means that IIN and also, by the definition of δN-1, λ(I)δN-1<2λ(IN). Write I=[y-r,y+r]. IIN tells us y-rxN+rN and y+rxN-rN, and λ(I)<2λ(IN) tells us 2r<4rN, hence


showing that


This is true for each xEj=1Ij, which means that


Because λ(j=MJj)0 as M, this yields


But Ej=1nIj is an increasing sequence of sets tending to Ej=1Ij, therefore


so there is some n such that λ*(Ej=1nIj)<ϵ and then I1,,In satisfy the claim. ∎