continuous
function can
map a set with outer measure zero onto a set with outer measure one.
Solution:
Choose
E
=
C
to be the
Cantor set
contained in [0
,
1] and let
f
(
x
) be the
continuous Cantor Lebesgue function de
fi
ned on [0
,
1] (see book p.
35)
.
We know that when
restricted to
C, f
(
x
) is still a
continuous
function on
C.
Moreover, one can easily see that
f
(
C
) =
f
([0
,
1]) = [0
,
1] (for example, we have
f
¡£
1
3
,
2
3
¤¢
=
f
¡©
1
3
,
2
3
ª¢
,
etc.).
¤
3. (10 points) Let
E
1
and
E
2
be two subsets of
R
n
such that
E
1
⊂
E
2
and
E
2
−
E
1
is
countable. Show that

E
1

e
=

E
2

e
.
Solution:
Clearly we have

E
1

e
≤

E
2

e
.
Also

E
2

e
≤

E
1

e
+

E
2
−
E
1

e
=

E
1

e
which implies

E
1

e
=

E
2

e
.
¤
4. (10 points) Find a continuous function
f
(
x
) de
fi
ned on [0
,
1] such that
f
(
x
) is di
ff
erentiable
on a subset
E
⊂
[0
,
1] with

E

e
= 1 and
f
0
(
x
) = 0 for all
x
∈
E,
but
f
(
x
) is not a constant
function.
1
Solution
: Let
f
(
x
) be the Cantor Lebesgue function de
fi
ned on [0
,
1] as given in p. 35 of
the book. We know
f
(
x
) is di
ff
erentiable on the open set
O
=
O
1
∪
O
2
∪
O
3
∪
· · ·
,
where
O
1
=
μ
1
3
,
2
3
¶
,
O
2
=
μ
1
9
,
2
9
¶
∪
μ
7
9
,
8
9
¶
,
O
3
=
μ
1
27
,
2
27
¶
∪
· · ·
,
O
4
=
· · ·
The total length of these open intervals is given by
1
3
"
1 +
2
3
+
μ
2
3
¶
2
+
μ
2
3
¶
3
+
· · ·
#
= 1
.
¤
The example below shows you how to obtain a set which is measurable, but
not
Borel
measurable.
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 Fall '08
 Akhmedov,A
 Continuous function, Lebesgue measure, Borel, Cantor Lebesgue, Cantor Lebesgue function