a limit theorem for the hurwitz zeta-function in the space of analytic ...

76
Proc. Sci. Seminar
Faculty of Physics and Mathematics, Šiauliai University,
5, 2002, 76–89
A LIMIT THEOREM
FOR THE HURWITZ ZETA-FUNCTION
IN THE SPACE OF ANALYTIC FUNCTIONS
Renata MACAITIENĖ
Faculty of Physics and Mathematics, Šiauliai University, Vytauto 84, 5400, Šiauliai,
Lithuania.
e-mail: renata.m@centras.lt
Abstract. In this paper a limit theorem for the Hurwitz zeta-function in the space of
analytic functions is proved.
Key words: Hurwitz zeta-function, probability measure, weak convergence.
1. Introduction
Let s = σ + it be a complex variable. The Hurwitz zeta-function ζ(s, α) is
defined, for σ > 1, by
∞∑ 1
ζ(s, α) =
(m + α) s .
m=1
and by analytic continuation elsewhere. Here 0 < α 1.
Denote by B(S) the class of Borel sets of the space S, and let, for T > 0,
ν T (...) = 1 meas{t ∈ [0, T ], ...},
T
where meas{A} stands for the Lebesgue measure of the set A, and in place of dots
some condition satisfied by t is to be written.
For value-distribution of zeta-functions probabilistic methods can be applied.
Denote by ζ(s) the Riemann zeta-function, and let C be the complex plane. Then
in [3] the following statement was proved.
Theorem 1. Let σ > 1 2
. Then there exists a probability measure on (C, B(C)) such
that the measure
ν T (ζ(σ + it) ∈ A), A ∈ B(C),
converges weakly to P as T → ∞.
1
Let D = {s ∈ C :
2
< σ < 1}, and let H(D) be the space of analytic on
D functions equipped with the topology of uniform convergence on compact sets.
Define probability measures
P T (A) = ν τ (τ ∈ [0; T ] : ζ(s + iτ, α), ∈ A),
A ∈ B(H(D)).

R. Macaitienė 77
Denote by γ the unit circle on C, i. e. γ = {s ∈ C : |s| = 1}, and let
Ω = ∏ m
γ m ,
where γ m = γ for each m = 0, 1, . . .. With the product topology and pointwise multiplication
Ω is a compact Abelian topological group. Then there exists the probability
Haar measure m H on (Ω, B(Ω)). This yields a probability space (Ω, B(Ω), m H ).
Denote by ω(m) the projection of ω ∈ Ω to the coordinate space γ m .
Let ζ(s, ω, α) be the H(D)-valued random element on (Ω, B(Ω), m H ) given by
the formula
∞∑ ω(m)
ζ(s, ω, α) =
(m + α) s .
m=1
Denote by P ζ the distribution of the random element ζ(s, ω, α), i. e.
P ζ (A) = m H (ω ∈ Ω : ζ(s, ω, α) ∈ A),
A ∈ B(H(D)).
The aim of this note is to prove the following statement. Suppose α is a transcendental
number.
Theorem 2. The probability measure P T converges weakly to P ζ as T → ∞.
2. Auxiliary results
Let
p n (s, α) =
n∑
m=1
1
(m + α) s
be an arbitrary Dirichlet polynomial, D denote some open subset of C, and let α is
a transcendental number. We define the probability measure
P T,pn (A) = ν τ T (p n (s + iτ, α) ∈ A),
A ∈ B(H(D)).
Lemma 1. There exists a probability measure P pn on (H(D), B(H(D))) such that
P T,pn converges weakly to P pn as T → ∞.
Proof. Let
Ω m =
n∏
γ m , γ m = γ.
m=0
Let us define the function h : Ω n → H(D) by the formula
h(x 0 , . . . , x n ) =
n∑
m=0
1
(m + α) s 1
x m
, (x 0 , . . . , x n ) ∈ Ω n .

78 A limit theorem for the Hurwitz zeta-function...
The function h is continuous on Ω n , and
p n (s + iτ, α) = h(α iτ , (1 + α) iτ , . . . , (n + α) iτ ). (1)
Now we define the probability measure
Q T (A) = νT τ ((α iτ , (1 + α) iτ , . . . , (n + α) iτ ) ∈ A)
on (Ω n , B(Ω n )). The Fourier transform g T (k 0 , . . . , k n ), k j ∈ Z, j = 1, . . . , n, of Q T
is given by the formula
∫
g T (k 0 , . . . , k n ) =
Ω
x k 0
0 , . . . , xkn n dQ T = 1 T
∫ T
∫ T
0
n∏
(j + α) ikmτ dτ
j=0
= 1 exp ( n∑
iτ k m log(m + α) ) dτ
T
0
m=0
⎧
1 if (k 0 , . . . , k n ) = (0, . . . , 0),
⎪⎨
( n∑
)
exp iT k m log(m+α)
=
m=0
n∑
if (k 0 , . . . , k n ) ≠ (0, . . . , 0).
⎪⎩ iT k m log(m+α)
m=0
(2)
Since α is transcendental, log(m + α), m = 0, 1, . . . , n, are linearly independent
over the field of rational numbers, whence we find that
g T (k 0 , . . . , k n ) =
{
1 if (k0 , . . . , k n ) = (0, . . . , 0),
0 if (k 0 , . . . , k n ) ≠ (0, . . . , 0)
as T → ∞. Therefore, the measure Q T converges weakly to the Haar measure m nH
on (Ω, B(Ω)) as T → ∞. Taking into account the continuity of the function h and
the formula (1), and applying Theorem 5.1 from [1], we obtain that the probability
measure P T,pn converges weakly to the measure m nH h −1 as T → ∞. The theorem
is proved.
Let g(m), m ∈ N, be an arithmetic function, |g(m)| = 1,
p n (s, g, α) =
n∑
m=1
g(m)
(m + α) s ,
and
˜P T,pn (A) = ν τ T
(
pn (s + iτ, g, α) ∈ A ) , A ∈ B(H(D)).
Lemma 2. The probability measures P T,pn and ˜P T,pn both converge weakly to the
same measure as T → ∞.

R. Macaitienė 79
Proof. Let the function h 1 : Ω n → Ω n be defined by the formula
h 1 (x 0 , . . . , x n ) = (x 0 e −iθ 0
, . . . , x n e −iθ n
),
where θ m = arg g(m), m = 0, . . . , n. By Lemma 1 the probability measures P n,pn
and ˜P n,pn converges weakly to the measures m nH h −1 and m nH˜h−1 , respectively,
where the function ˜h is similar to h. It is easy to see that
˜h(x0 , . . . , x n ) =
Consequently,
n∑
m=0
g(m)
(m + α) s x m
=
n∑
m=0
e iθm
(m + α)x m
= h(h 1 (x 0 , . . . , x n )).
m nH˜h−1 = m nH (h(h 1 )) −1 = m nH (h −1
1 )h−1 . (3)
Since the Haar measure m nH is invariant with respect to translation by points in Ω,
we have that m nH h −1
1 = m nH . It follows from this and from (3) that m nH˜h−1 =
m nH h −1
1 . This proves the lemma.
3. Aplication of the ergodic theory
Denote by D a compact Abelian topological group. Let m D be the Haar measure
on (D, B(D)). A one-parameter family {ϕ τ : τ ∈ R} of transformations on D is
called a one-parameter group of transformations if for τ, τ 1 , τ 2 ∈ R, g ∈ D, we have
ϕ τ1 +τ 2
(g) = ϕ τ1 (ϕ τ2 (g)) and ϕ −τ (g) = ϕ −1
τ (g).
Let a τ = {(m + α) −iτ : 0 < α < 1}, τ ∈ R, m ∈ N 0 , N 0 = N ∪ {0}. Then
{a τ : τ ∈ R} is a one-parameter group. We define the one-parameter family
{ϕ τ : τ ∈ R} of transformations on Ω taking ϕ τ (ω) = a τ ω, ω ∈ Ω. Then we have
that {ϕ τ : τ ∈ R} is a one-parameter group of measurable transformations on Ω.
Lemma 3. The one-parameter group {ϕ τ : τ ∈ R} is ergodic.
Proof is similar to that from [3].
Denote by EX the mean of the random element X.
Lemma 4. Let a process X(τ, ω) be ergodic, E|X(τ, ω)| < ∞ and let sample paths
be integrable almost surely in the Riemann sense over every finite interval. Then
almost surely.
Proof can be found in [2].
∫
1
T
lim X(τ, ω)dτ = EX(0, ω)
T →∞ T
0

80 A limit theorem for the Hurwitz zeta-function...
Lemma 5. Let T → ∞ and 1 2
< σ < 1. Then
∫ T
0
|ζ(σ + it, ω, α)| 2 dt = BT
for almost all ω ∈ Ω.
Proof. Let
Then
Denote
ζ m (σ, ω, α) =
ζ(σ, ω, α) =
ω(m)
(m + α) σ , m ∈ N 0.
∞∑
ζ m (σ, ω, α).
m=0
ˆζ(σ, ω, α) = |ζ(σ, ω, α)| 2 .
Since
E|ζ m (σ, ω, α)| 2 =
1
(m + α) 2σ ,
and the random variables ζ m (σ, ω, α) are pairwise ortogonal, we find that
It is obvious that
E|ˆζ(σ, ω, α)| =
∞∑
E|ζ m (σ, ω, α)| 2 =
m=0
∞∑
m=1
1
< ∞. (4)
(m + α)
2σ
ˆζ(σ, ϕ τ (ω), α) = |ζ(σ, a τ ω, α)| 2 = |ζ(σ + iτ, ω, α)| 2 . (5)
Since the Haar measure is invariant, the equality m H (ϕ τ (A)) = m H (A) is valid
for each A ∈ B(Ω) and every τ ∈ R. Therefore |ζ(σ + iτ, ω, α)| 2 is a strongly
stationary process. It is also an ergodic process. In fact, let A be an invariant set
of |ζ(σ + iτ, ω, α)| 2 , i. e.
Q(A∆A u ) = 0. (6)
We have that
A ′
A ′ u
def
= {ω ∈ Ω : |ζ(σ + iτ, ω, α)| 2 ∈ A} = {ω ∈ Ω : |ζ(σ, a τ ω, α)| 2 ∈ A},
def
= {ω ∈ Ω : |ζ(σ + iτ, ω, α)| 2 ∈ A u } = {ω ∈ Ω : |ζ(σ + iτ + iu, ω, α)| 2 ∈ A}
= {ω ∈ Ω : |ζ(σ + iτ, a u ω, α)| 2 ∈ A}.
Therefore A ′ u = ϕ u (A ′ ) and (A∆A u ) ′ = A ′ ∆A ′ u. From this and from (6) we deduce
that m H (A ′ ∆A ′ u) = m H ((A∆A u ) ′ ) = Q(A∆A u ), that is A ′ is an invariant set with
respect to ϕ τ . But, by Lemma 3, the group {ϕ τ : τ ∈ R} is ergodic. Therefore

R. Macaitienė 81
m(A ′ ) = 0 or m(A ′ ) = 1. Hence it follows that Q(A) = 0 or Q(A) = 1, i. e. the
process |ζ(σ + iτ, ω, α)| 2 is ergodic.
Observing that ˆζ(σ, ϕ τ (ω), α) 0 and using (4), (5) we deduce from Lemma 4
that
∫
1
T
∫
1
T
lim ˆζ(σ, ϕ τ (ω), α)dt = lim |ζ(σ + iτ, ω, α)| 2 dτ = E
T →∞ T
T →∞ T
ˆζ(σ, ω, α)
0
0
for almost all ω ∈ Ω. From this by (4) we obtain the assertion of the lemma.
4. Approximation by mean of the function ζ(s, α)
Let σ 1 > 1 2
. We define the function
l n (s, α) = s σ 1
Γ ( s
σ 1
)
(n + α) s , n ∈ N,
in the strip 1 2 < σ < 1. Moreover, let for σ 1 > 1 2
ζ 2,n (s, α) = 1
2πi
∫
σ 1+i∞
σ 1−i∞
ζ(s + z, α)l n (z, α) dz
z .
Since Γ(σ + it) = Be −c 1|t| , uniformly in σ, |σ| c 1 , the integral for ζ 2,n (s, α) exists.
Lemma 6. Let a and b be positive numbers. Then the formula
is valid.
Proof can be found in [4].
1
2πi
∫
b+i∞
b−i∞
Γ(s)a −s ds = e −a
Lemma 7. Let K be a compact subset of the strip σ > 1 2 . Then
lim
lim
∫ T
n→∞ T →∞
0
sup |ζ(σ + iτ, α) − ζ 2,n (s + iτ, α)|dτ = 0.
s∈K

82 A limit theorem for the Hurwitz zeta-function...
Proof. By choice of σ 1 and σ we have that σ+σ 1 > 1. Consequently, the function
ζ(s + z, α) for Rez = σ 1 , is represented by an absolutely convergent Dirichlet series
Let
Consider the series
Since
ζ(s + z, α) =
a n (m, α) = 1
2πi
a n (m, α) = B(m + α) −σ 1
∞∑
m=1
∫ ∞
−∞
∞∑
m=1
1
(m + α) s+z .
σ 1+i∞ ∫
σ 1 −i∞
l n (s, α)ds
s(m + α) s .
a n (m, α)
(m + α) s . (7)
|l n (σ 1 + it, α)|dt = B(m + α) −σ 1
,
the series (7) converges absolutely for σ > 1 2
. Thus, interchanging sum and integral
in the definition of ζ 2,n , we find
By Lemma 6
ζ 2,n (s, α) =
∞∑
m=1
{
a n (m, α) = exp −
Therefore the equality (8) can be writen as
ζ 2,n (s, α) =
∞∑
m=1
{
1
(m + α) s exp −
a n (m, α)
(m + α) s . (8)
( ) σ1
}
(m + α)
.
(n + α)
( ) σ1
}
(m + α)
,
(n + α)
the series being absolutely convergent for σ > 1 2 .
Now we change the contour in the integral for ζ 2,n (s, α). The integrand has
simple poles at z = 0 and z = 1 − s. Let σ belong to σ ∈ [ 1 2
+ ε, 1 − ε] when s ∈ K.
We put
Then by the residue theorem
ζ 2,n (s, α) = 1
2πi
∫
σ 2−σ+i∞
σ 2−σ−i∞
σ 2 = 1 2 + ε 2 .
ζ(s + z, α)l n (z, α) dz
z + ζ(s, α) + l n(1 − s, α)
. (9)
1 − s

R. Macaitienė 83
1
Let L be a simple closed contour lying in {s ∈ C :
2
< σ < 1} and enclosing
the set K, and let δ denote the distance of L from the set K. Then by the Cauchy
formula we have
sup |ζ(s + iτ, α) − ζ 2,n (s + iτ, α)| 1 ∫
|ζ(z + iτ, α) − ζ 2,n (z + iτ, α)||dz|.
s∈K
2πδ
Therefore we obtain that, for sufficiently large T ,
∫
1
T
sup |ζ(s + iτ, α) − ζ 2,n (s + iτ, α)|dτ
T s∈K
0
= B T δ
∫
L
= B|L|
T δ
∫
|dz|
2T
0
+ B|L|
T δ
L
|ζ(Rez + iτ, α) − ζ 2,n (s + iτ, α)|dτ + B|L|
T δ
sup
∫2T
σ,s∈L
0
The contour L can be chosen so that the inequalities
|ζ(s + it, α) − ζ 2,n (s + it, α)|dt.
σ 1 2 + 3ε
4 , δ ε 4
should hold. Then by (9) we have that for such σ
ζ(s + it, α) − ζ 2,n (s + it, α) = B
∫ ∞
−∞
|ζ(σ 2 + it + iτ, α)||l n (σ 2 − σ + iτ, α)|dτ
+ B|l n(1 − σ − it, α)|
.
|1 − σ − it|
Hence, for the same σ, in virtue of the properties of l n (s, α) and ζ(s, α), we find
that
1
T
∫2T
0
= B
|ζ(σ + it, α) − ζ 2,n (σ + it, α)|dt
∫ ∞
−∞
+ B T
∫2T
0
|l n (σ 2 − σ + iτ, α)| 1 T
|l n (1 − σ − it, α)|
dt.
|1 − σ − it|
|τ|+2T
∫
−|τ|
|ζ(σ 2 + it, α)|dtdτ
(11)

84 A limit theorem for the Hurwitz zeta-function...
It is known that
∫ T
Now it follows from (11) and (12) that
1
T sup
∫2T
σ,s∈L
0
= B sup
0
|ζ(σ 2 + it, α)| 2 dt = BT. (12)
|ζ(σ + it, α) − ζ 2,n (σ + it, α)|dt
∫ ∞
σ,s∈L
−∞
+ B T sup
∫2T
σ,s∈L
0
= B sup
σ∈[− 1 2 + 3ε
|l n (σ 2 − σ + iτ, α)| ( 2 + 2|τ| )
dτ
T
|l n (1 − σ − it, α)|
dt
|1 − σ − it|
∫
2 ,− ε 4 ] ∞
−∞
|l n (σ + it, α)|(1 + |τ|)dτ + o(1)
as T → ∞. From the definition of the function l n (s, α) we deduce that
∫
lim sup |l n (σ + it, α)|(1 + |τ|)dτ = 0.
∞
n→∞
σ∈[− 1 2 + 3ε
2 ,− ε 4 ] −∞
This equality together with (13) ir (10) gives the assertion of the theorem.
(13)
5. Limit theorem in the space of analytic functions for the absolutely
convergent series
Let for ω ∈ Ω,
ζ 2,n (s, ω, α) =
∞∑
m=1
{
ω(m)
(m + α) s exp −
( ) σ1
}
(m + α)
.
(n + α)
Since |ω(m)| = 1, the series for ζ 2,n (s, ω, α) converges absolutely on the half-plane
σ > 1 2
. We define two probability measures on (H(D), B(H(D)))
P 1 T,n(A) = ν τ T (ζ 2,n (s + iτ, α) ∈ A),
Q 1 T,n(A) = ν τ T (ζ 2,n (s + iτ, ω, α) ∈ A).
Lemma 8. There exists a probability measure P 1 n on (H(D), B(H(D))) such that
both measures P 1 T,n and Q1 T,n converge weakly to P 1 n as T → ∞.
Proof is similar to that for the case of the Riemann zeta-function and uses
Lemma 2.

R. Macaitienė 85
6. Proof of Theorem 2
Let Ω 1 be a subset of Ω such that for ω 1 ∈ Ω 1 the series
∞∑
m=1
ω 1 (m)
(m + α) s
converges uniformly on compact subsets of D and, let for σ > 1 2
, the estimate
be valid. By Lemma 5 we have that m H (Ω 1 ) = 1.
Lemma 9. Let K be a compact subsed of D. Then
for ω 1 ∈ Ω 1 .
lim
lim
n→∞ T →∞
1
T
∫ T
0
∫ T
0
|ζ(σ + it, ω 1 , α)| 2 dt = BT (14)
sup |ζ(s + iτ, ω 1 , α) − ζ n (s + iτ, ω 1 , α)|dτ = 0,
s∈K
Proof. The lemma is proved similarly to Lemma 7, since by (14) and the Cauchy
inequality the estimate
∫ T
follows.
Let
0
|ζ(σ + it, ω 1 , α)|dt = BT
Q T (A) = ν τ T (ζ(s + iτ, ω 1 , α) ∈ A),
A ∈ B(H(D)).
Lemma 10. There exists a probability measure P 1 on (H(D), B(H(D))) such that
the measures P T and Q T both converge weakly to P 1 as T → ∞.
Proof. By Lemma 8 both the measures PT,n 1 and Q1 T,n converge weakly to the
same measure Pn 1 as T → ∞. First we will prove that the family of probability
measures {Pn} 1 is tight.
Let θ be a random variable, uniformly distributed in the interval [0, 1] and
Then by Lemma 8 we have that
X T,n (s, α) = ζ 2,n (s + iT θ, α).
X T,n
D
−→ Xn , T → ∞, (15)

86 A limit theorem for the Hurwitz zeta-function...
where X n is an H(D)-valued random element having the distribution P 1 n.
Let {K l } be a sequence of compact subsets of D such that
∞⋃
K l = D,
l=1
K l ⊂ K l+1 , and if K is a compact of D, then K ⊆ K l for some l. Define
ρ(f, g) =
∞∑
l=1
2 −l ρ l (f, g)
, f, g ∈ H(D),
1 + ρ l (f, g)
where ρ l (f, g) = sup
s∈K l
|f(s) − g(s)|. Then ρ is a metric on H(D) which induces its
topology.
Since the series for ζ 2,n (s, α) is absolutely convergent in D, we have
sup
lim
n1 T →∞
1
T
∫ T
0
sup |ζ 2,n (s + iτ, α)|dτ R l < ∞. (16)
s∈K l
Let ε > 0 and M l = R l2 l
ε
. Then by the Chebyshev inequality
P ( )
sup |X T,n (s, α)| > M 1
l
s∈K l
T M l
∫T
0
sup |ζ 2,n (s + iτ, α)|dτ,
s∈K l
and, consequently,
Thus we find that
and, for all n 1,
lim P( )
sup |X T,n (s, α)| > M ε l
T →∞ s∈K l
2 l .
P ( )
sup |X n (s, α)| > M ε l
s∈K l
2 l ,
P ( X n (s, α) ∈ H ε
)
1 − ε
or, since P 1 n is the distribution of X n ,
P 1 n(H ε ) 1 − ε,
where H ε = {h ∈ H(D) : sup |h(s)| M l , l 1}. The set H ε is compact.
s∈K l
Therefore the family of the measures {Pn} 1 is tight, and by the Prokhorov theorem
it is relatively compact.

R. Macaitienė 87
Aplying the Chebyshev inequality once more and Lemma 7, we deduce that for
every ε > 0
lim lim
(
ρ(ζ(s + iτ, α), ζ2,n (s + iτ, α)) ε )
Let
n→∞ T →∞ ντ T
∫ T
lim
lim
n→∞ T →∞
1
εT
0
ρ(ζ(s + iτ, α), ζ 2,n (s + iτ, α))dτ = 0.
Y T (s, α) = ζ(s + iT θ, α).
Then the relation (17) can be written in the form
lim
n→∞
(17)
lim P(ρ(X T,n(s, α), Y T (s, α)) ε) = 0. (18)
T →∞
Let {P 1 n ′} bet a subsequence of {P 1 n} which converges weakly to some measure
P 1 . Since the space H(D) is separable, by (15) we have that
X n ′
D
−→ P 1 , n ′ → ∞.
From this and the relations (15) and (18), we deduce that
Y T
D
−→ P 1 , T → ∞. (19)
This relation is equivalent to the weak convergence of P T to P 1 . From (19) in view
of the relative compactness of the family {Pn} 1 and of the assertion (The relation
P n ⇒ P is true if and only if every subsequence P n ′ contains another subsequence
P n ′′ such that P n ′′ ⇒ P ) we get
X n
D
−→ P 1 , n → ∞. (20)
and
Repeating the analogous reasoning for the random elements
˜X T,n (s, ω 1 , α) = ζ 2,n (s + iT θ, ω 1 , α)
Ỹ T (s, ω 1 , α) = ζ(s + iT θ, ω 1 , α)
and applying Lemma 9 and (20), we obtain that the measure Q T converges weakly
to P 1 as T → ∞. The lemma is proved.
Proof of Theorem 2. Lemma 10 asserts that the measures P T and Q T converge
weakly to some measure P 1 as T → ∞ simultaneously. It remains to prove that
P 1 = P ζ .
Let A ∈ B(H(D)) be a continuity set of P 1 . Then by Lemma 10 we have that
lim
T →∞ ντ T (ζ(s + iτ, ω 1 , α) ∈ A) = P 1 (A). (21)

88 A limit theorem for the Hurwitz zeta-function...
Let us fix the set A and define the random variable θ on (Ω, B(Ω), m H ) by the
formula
{
1 if ζ(s, ω, α) ∈ A,
θ(ω) =
0 if ζ(s, ω, α) /∈ A.
It is easy to verify that
∫
E(θ) =
Ω
θdm H = m H (ω : ζ(s, ω, α) ∈ A) = P ζ < ∞. (22)
Taking into account Lemma 3 an reasoning similarly as in the proof of Lemma 5,
we obtain that θ(ϕ τ (ω)) is an ergodic process. Therefore in view of Lemma 4
lim
∫ T
T →∞
0
θ(ϕ τ (ω))dτ = E(θ) (23)
for almost all ω ∈ Ω. But from the definitions of the random variable θ and of the
one-parameter group {ϕ τ : τ ∈ R} it follows that
∫ T
0
θ(ϕ τ (ω))dτ = ν τ T (ζ(s + iτ, ω, α) ∈ A).
This, (22) and (23) give us
for almost all ω. Thus, in view of (21)
( )
lim
T →∞ ντ T ζ(s + iτ, ω, α) ∈ A = Pζ (A)
P 1 (A) = P ζ (A)
for any continuity set A of the measure P 1 . Since the continuity sets constitute the
determining class, we have that
P 1 (A) = P ζ (A)
for all A ∈ B(H(D)).
The theorem is proved.

R. Macaitienė 89
REFERENCES
[1] P. Billingsley, Convergence of Probability Measures, John Wiley, New York (1968).
[2] H. Cramér, M. R. Leadbeeter, Stationary and Related Stochastic Processes, Wiley, New York
(1967).
[3] A. Laurinčikas, Limit Theorems for the Riemann zeta-function, Kluwer, Dordrecht, Boston,
London (1996).
[4] E. C. Titchmarsh, The Theory of Functions, Oxford University Press (1939).
[5] B. V. Shabat, Introduction to Complex analysis, Nauka, Moscow (1969).
Ribinė teorema Hurwitz’o dzeta funkcijai analiziniu ‘
funkciju ‘
erdvėje
R. Macaitienė
Straipsnyje i ‘
rodoma ribinė teorema Hurvitz’o dzeta funkcijai analiziniu ‘
funkciju ‘
erdvėje.
Rankraštis gautas
2002 07 04