Problem. Prove that if f : [0, 1] → R is a function, H ⊂ [0, 1] such that f 0 (x) = 0 for every x ∈ H, then
λ(f (H)) = 0.
Solution. Let ε > 0 be arbitrary, we will show that λ∗ (f (H)) < 8ε.
(y)
If x ∈ H then by the definition of differentiability there exists a δx > 0 such that | f (x)−f
| < ε whenever
x−y
y ∈ (x − δx , x + δx ) and y 6= x. Let us denote this
interval
by
I
.
Thus,
for
every
y
∈
I
with
y 6= x we have
x
x
S
|f (x) − f (y)|
<
ε|x
−
y|
<
εδ
.
Clearly,
H
⊂
I
,
so
there
exist
countably
many
intervals
Ix1 , Ix2 , . . .
x
x∈H x
S∞
S∞
with H ⊂ n=1 Ixn . The union n=1 Ixn can be expressed as a countable disjoint union of open intervals
J1 , J2 , . . . .
The following easy claim can be proved by induction.
Sn
Claim 1. Suppose that (a1 , b1 ), . . . , (an , bn ) is a finite sequence of intervals
such that [a, b] ⊂ i=1 (ai , bi ).
Sn
Then we can find a subsequence an1 , bn1 , . . . , ank , bnk such that [a, b] ⊂ j=1 (anj , bnj ) and they form a chain,
i. e. an1 < an2 < bn1 < an3 < bn2 < . . . and each intersects [a, b].
Claim 2. If y, z ∈ Jj for some j then |f (y) − f (z)| < 4ελ(Jj ).
Proof of Claim 2. We know that Jj is the union of intervals of the form Ixn so the (compact) interval
[y, z] can be covered by finitely many such intervals. By the first claim we can also select a chain Ixn1 ,
Ixn2 , . . . from these intervals. Let us use the notation yj = xnj . If [y, z] is covered by only one interval Iy1
then of course |f (y) − f (z)| ≤ |f (y) − f (y1 )| + |f (y1 ) − f (z)| ≤ 2εδy1 ≤ 2ελ(Jj ).
If y, z is covered by several intervals, chose points w1 , w2 , . . . , wn−1 from the intersections of Iy1 and Iy2 ,
Iy2 and Iy3 ,. . . ,Iyn−1 and Iyn , respectively. Then
|f (y) − f (z)| ≤ |f (y) − f (y1 )| + |f (y1 ) − f (w1 )| + |f (w1 ) − f (y2 )| + · · · + |f (yn ) − f (z)| ≤
ε(2δy1 + δy2 + · · · + δyn−1 + 2δyn )
and the last sum is ≤ 4ελ(Jj ) as by the first claim every point is covered at most twice. This proves the
claim.
S∞
Finally, choosing a point pj from each Jj and using Claim 2 we get that f (H) ⊂ j=1 (f (pj ) −
P
∞
4ελ(Jj ), f (pj ) + 4ελ(Jj )), so λ∗ (f (H)) ≤ 8ε( j=1 λ(Jj )) ≤ 8ε, since the Jj intervals are disjoint and
contained in [0, 1] so their total measure is at most 1.