an elementary proof of lebesgue's differentiation theorem

Since E ⊆∞⋃(a n , b n )n=1∞∑(b n − a n ) ≥ ε.n=1Now for all (a n , b n ) choose a closed subinterval [a ′ n, b ′ n] such that c > 1, c ∈ R,∞∑b ′ n − a ′ n = (b n − a n )/c. Thus (b ′ n − a ′ n) ≥ ε/c.n=1Let n be a fixed positive integer. For each x in [a ′ n, b ′ n] there exists J x inI such that x ∈ J x ⊆ (a n , b n ). So {J x : x ∈ [a ′ n, b ′ n]} is an open cover of[a ′ n, b ′ n]. Since [a ′ n, b ′ n] closed and bounded by Heine-Borel theorem it is compact.So an open cover {J x : x ∈ [a ′ n, b ′ n]} of [a ′ n, b ′ n] can be reduced to a finitesubcovering {J 1 , J 2 , . . . J p } such that [a ′ n, b ′ n] ⊆ ⋃ pk=1 J k.Furthermore by discarding some of the intervals, we may assume that nointerval in {J k } p k=1is a subset of the union of the remaining intervals in{J k } p k=1 . Thus each J i contains a point x i which does not belong to ⋃ k≠i J k,and renumbering J k s, if necessary, we can assume that x 1 < x 2 < . . . < x p .Therefore both {J 1 , J 3 , J 5 , . . .} and {J 2 , J 4 , J 6 , . . .} are finite disjoint subcollectionsof I.It is obvious that either∑ ( ∑p )|J 2k−1 | ≥ |J k | /2 orkk=1∑ ( ∑p )|J 2k | ≥ |J k | /2.kk=1Depending on which one of the inequalities holds, we have found a finitedisjoint subcollection I n of I such that∑ ( ∑p )|I| ≥ |J k | /2 ≥ (b ′ n − a ′ n)/2.I∈I nk=1Hence for each positive integer n there exist a finite disjoint subcollection I nof I each of whose open intervals is a subset of (a n , b n ) such that∑I∈I n|I| ≥ (b ′ n − a ′ n)/2.Summing both sides of the inequality, we have∞∑ ∑ ( ∑ ∞ )|I| ≥ (b ′ n − a ′ n) /2 ≥ ε/2c.n=1 I∈I n n=13