Measure, Integration & Real Analysis, 2021a

114 Chapter 4 Differentiation
Proof
We use the following fact in our construction:
4.26 Suppose G is a nonempty open subset of R. Then there exists a closed set
F ⊂ G \ Q such that |F| > 0.
To prove 4.26, let J be a closed interval contained in G such that 0 < |J |. Let
r 1 , r 2 ,...be a list of all the rational numbers. Let
∞⋃ (
F = J \ r k − |J |
2 k+2 , r k + |J | )
2 k+2 .
k=1
Then F is a closed subset of R and F ⊂ J \ Q ⊂ G \ Q. Also, |J \ F| ≤ 1 2 |J |
because J \ F ⊂ ⋃ (
)
∞
k=1
r k − |J | , r
2 k+2 k + |J | . Thus
2 k+2
|F| = |J |−|J \ F| ≥ 1 2
|J | > 0,
completing the proof of 4.26.
To construct the set E with the desired properties, let I 1 , I 2 ,... be a sequence
consisting of all nonempty bounded open intervals of R with rational endpoints. Let
F 0 = ̂F 0 = ∅, and inductively construct sequences F 1 , F 2 ,...and ̂F 1 , ̂F 2 ,...of closed
subsets of R as follows: Suppose n ∈ Z + and F 0 ,...,F n−1 and ̂F 0 ,...,̂F n−1 have
been chosen as closed sets that contain no rational numbers. Thus
I n \ ( ̂F 0 ∪ ...∪ ̂F n−1 )
is a nonempty open set (nonempty because it contains all rational numbers in I n ).
Applying 4.26 to the open set above, we see that there is a closed set F n contained in
the set above such that F n contains no rational numbers and |F n | > 0. Applying 4.26
again, but this time to the open set
I n \ (F 0 ∪ ...∪ F n ),
which is nonempty because it contains all rational numbers in I n , we see that there is
a closed set ̂F n contained in the set above such that ̂F n contains no rational numbers
and |̂F n | > 0.
Now let
∞⋃
E = F k .
k=1
Our construction implies that F k ∩ ̂F n = ∅ for all k, n ∈ Z + . Thus E ∩ ̂F n = ∅ for
all n ∈ Z + . Hence ̂F n ⊂ I n \ E for all n ∈ Z + .
Suppose I is a nonempty bounded open interval. Then I n ⊂ I for some n ∈ Z + .
Thus
0 < |F n |≤|E ∩ I n |≤|E ∩ I|.
Also,
completing the proof.
|E ∩ I| = |I|−|I \ E| ≤|I|−|I n \ E| ≤|I|−|̂F n | < |I|,
Measure, Integration & Real Analysis, by Sheldon Axler