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0
f(z)ν(
d
z)
=
lim
a
→
f(a)ν(I(a))
−
lim
b
→−∞
f(b)ν(I(b))
0
−
−∞
0
−
∂
1
f(z)ν(I(z))
d
z,
−∞
and since
f
is bounded,
f(z)ν(
d
z)
=
f(
0
)
lim
a
(ν (I(a))
+
ν (I(
−
a)))
+
∂
1
f(z)
sgn
(z)ν (I(z))
d
z.
→
0
+
R
R
Abusing notation, we write
ν(
R
)
:=
lim
(ν (I(a))
+
ν (I(
−
a))) .
a
→
0
+
With
f
vanishing on a neighborhood of 0, we therefore find
f(
0
)ν(
0.
For the multidimensional case, i.e. for
d>
1, we use the Lévy measure of
X
I
which is given by
R
)
=
ν
,
∈
B
R
|
I
|
.
ν
I
(B)
d
:=
{
x
∈ R
:
(x
i
)
i
∈
I
∈
B
\{
0
}}
∀
B
We show by induction with respect to the dimension
d
that
f(z)ν(
d
z)
=
f(
0
,...,
0
)ν(
R
,...,
R
)
R
d
d
+
∂
i
f(
0
,...,z
i
,...,
0
)
sgn
(z
i
)ν
i
(I(z
i
))
d
z
i
R
i
=
1
sgn
(z
j
)ν
I
j
I(z
j
)
d
z
I
.
d
z
I
)
j
∂
I
f(
0
+
+
i
R
|
I
|=
i
I
1
<
···
<
I
i
∈
I
∈
I
i
=
2
With
f(
0
,...,
0
)ν(
0, the definition of the tail integrals and Theo-
rem
14.2.6
we then have the required result.
For the induction step
d
−
R
,...,
R
)
=
1
→
d
, using integration by parts and the induction
hypothesis, we obtain
f(z
,z
d
)ν(
d
z
,
d
z
d
)
f(z)ν(
d
z)
=
R
d
R
d
−
1
R
f(z
,
0
)ν(
d
z
,
=
R
)
d
−
1
R
∂
d
f(z
,z
d
)
sgn
(z
d
)ν
d
z
,I(z
d
)
d
z
d
+
d
−
R
1
R
=
f(
0
,...,
0
)ν(
R
,...,
R
)
d
−
1
+
∂
i
f(
0
,...,z
i
,...,
0
)
sgn
(z
i
)ν
i
(I(z
i
))
d
z
i
R
i
=
1
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