J Theor Probab DOI 10.1007/s10959-013-0526-8

Extensions of the Borel–Cantelli lemma in general measure spaces Xuejun Wang · Xinghui Wang · Xiaoqin Li · Shuhe Hu

Received: 23 April 2012 / Revised: 4 December 2012 © Springer Science+Business Media New York 2013

Abstract In this paper, an important bilateral inequality for a sequence of nonnegative measurable functions space (S, B S , μ) is obtained, and some sufficient   on a measure conditions for μ lim sup An n→∞

= μ(S) are given. In addition, a weighted version of

the Borel–Cantelli Lemma on the measure space is obtained. Our results generalize the corresponding ones for bounded random sequences to the case of unbounded measurable functions. Keywords

Borel–Cantelli lemma · Bilateral inequality · Measure space

Mathematics Subject Classification (2010)

60E15 · 60F20

1 Introduction A traditional form of the Borel–Cantelli Lemma states  that if A1 , A2 , · · · is a ∞ sequence of events on a probability space (, F, P) and n=1 P(An ) < ∞, then   P lim sup An n→∞

= 0; if A1 , A2 , · · · is a sequence of independent events and

Supported by the National Natural Science Foundation of China (11201001, 11171001), the Natural Science Foundation of Anhui Province (1208085QA03, 1308085QA03), Applied Teaching Model Curriculum of Anhui University (XJYYXKC04), Doctoral Research Start-up Funds Projects of Anhui University, Students Innovative Training Project of Anhui University (201310357004) and the Students Science Research Training Program of Anhui University (KYXL2012007, kyxl2013003). X. Wang · X. Wang · X. Li · S. Hu (B) School of Mathematical Science, Anhui University, Hefei 230601, People’s Republic of China e-mail: [email protected]

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J Theor Probab ∞ 

P(An ) = ∞,

(1.1)

n=1

  ∞  ∞  Ak . then P lim sup An = 1. Here, lim sup An = n→∞

n→∞

n=1 k=n

Many attempts have been made to weaken the independent conditions in the second part of the Borel–Cantelli Lemma, see for example, Erdös and Rényi [1], Lamperti [2], Kochen and Stone [3], Spitzer [4], Móri and Székely [5], Ortega and Wschebor [6], Petrov [7], Amghibech [8] and Yan [9]. For the conditional versions of generalized the Borel–Cantelli Lemma, one can refer to Prakasa Rao [10]. For more details about the Borel–Cantelli Lemma and its applications, one can refer to Chandra [11]. Petrov [12] gave a general result on the generalization of the second part of the Borel–Cantelli Lemma as the following theorem. Theorem 1.1 (cf. Petrov [12]) Let A1 , A2 , · · · be a sequence ofs events satisfying condition (1.1), H be an arbitrary real constant. Put  . 1≤i
Chandra [13] obtained the generalization of the second part of the Borel–Cantelli Lemma under weak dependence conditions, and the results of Chandra [13] are extensions of those of Petrov [7,12]. Xie [14] obtained a bilateral inequality on the Borel– Cantelli Lemma as follows. Theorem 1.2 (cf. Xie [14]) Let A1 , A2 , · · · be a sequence of eventssatisfying conn I Ai , Tn = dition (1.1), I Ai be the indicator function for the event Ai , Sn = i=1 Sn I(Sn >0) /E Sn . Then sup

  p 1 p 1 lim sup(E Tn ) 1− p ≤ P lim sup An ≤ inf lim inf (E Tn ) 1− p . (1.2)

p>0, p=1 n→∞

n→∞

p<0 n→∞

Hu et al. [15] pointed out that there were several mistakes in the proof of Theorem 1 and Example 2 in Xie [14], and they presented a corrected version of some results given therein. Recently, Xie [16] extended the result in Xie [14] to the case of nonnegative bounded random sequence. In this paper, we extend the results of Xie [14,16] for nonnegative bounded random sequences to the case of nonnegative unbounded measurable function sequences on a measure space (S, B S , μ) and give a new extension   of the Borel– Cantelli Lemma. In addition, we give some sufficient conditions for μ lim sup An n→∞

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J Theor Probab

μ(S). At last, we will present a weighted version of the Borel–Cantelli Lemma on a measure space. The following lemmas are needed. Lemma 1.1 Let p > 0. Assume that { f n (x), n ≥ 1} and {gn (x), n ≥ 1} are sequences of nonnegative measurable functions on a measure space (S, B S , μ). p Assume further that gn (x) ≤ g1 (x), lim gn (x) = 0 a.e. μ and S g1 (x)dμ(x) < n→∞

∞. Then, 



p

( f n (x) + gn (x)) p dμ(x) = lim sup

lim sup n→∞

f n (x)dμ(x).

n→∞

S

(1.3)

S

Proof If 0 < p ≤ 1, by Cr inequality, it follows that 



p

f n dμ ≤ S



p

( f n + gn ) p dμ ≤ S



f n dμ + S

p

gn dμ. S

We have by the Lebesgue dominated convergence theorem that 



p

f n dμ ≤ lim sup

lim sup n→∞

n→∞

S

 ( f n + gn ) p dμ ≤ lim sup n→∞

S

p

f n dμ, S

which implies (1.3). If p > 1, by Minkowski’s inequality, we can see that ⎛ ⎝



⎞1/ p p f n dμ⎠

⎛ ⎞1/ p ⎛ ⎞1/ p ⎛ ⎞1/ p    p p ≤ ⎝ ( f n + gn ) p dμ⎠ ≤ ⎝ f n dμ⎠ + ⎝ gn dμ⎠ .

S

S

S

S

It follows that by the Lebesgue dominated convergence theorem that ⎞1/ p ⎞1/ p ⎞1/ p ⎛ ⎛ ⎛    p p lim sup⎝ f n dμ⎠ ≤ lim sup⎝ ( f n + gn ) p dμ⎠ ≤ lim sup⎝ f n dμ⎠ , n→∞

n→∞

S

n→∞

S

S

thus ⎛ lim sup ⎝



n→∞

⎞1/ p p f n dμ⎠

⎞1/ p ⎛  = lim sup ⎝ ( f n + gn ) p dμ⎠ , n→∞

S

which implies (1.3). The proof is completed.

S

 

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Lemma 1.2 (Generalized Hölder’s inequality on measure space) Let f (x) and g(x) be measurable functions on a measure space (S, B S , μ). If 0 < p < 1, 1p + q1 = 1, then 

⎛ ⎞1 ⎛ ⎞1 p q   | f (x)g(x)|dμ(x) ≥ ⎝ | f (x)| p dμ(x)⎠ ⎝ |g(x)|q dμ(x)⎠ .

S

S

S

Proof Denote f = | f g| p , g = |g|− p . Since 0 < p < 1, inequality, we can see that 

⎛

| f | p dμ = S

⎛ =⎝ ⎛ =⎝





> 1, by Hölder’s

1 p

S

 S



S

⎞p ⎛ ⎞1− p  −p | f g|dμ⎠ ⎝ |g| 1− p dμ⎠ S

⎞p ⎛ ⎞1− p  | f g|dμ⎠ ⎝ |g|q dμ⎠ .

S

Note that q < 0, if

1 p

⎞p ⎛ ⎞1− p  1 f g dμ ≤ ⎝ ( f ) dμ⎠ ⎝ (g ) 1− p dμ⎠



S

(1.4)

S

|g|q dμ > 0, then by reorganizing the inequality above, we can

S

get



 | f g|dμ ≥ S

i.e. (1.4)

holds.

If S |g|q dμ = S



p S | f | dμ 

1− p q S |g| dμ

1 dμ |g|−q

1

= 0, then

p

⎛ =⎝



p

| f | dμ⎠ ⎝

S

1 |g|−q



p

⎞1 q

|g| dμ⎠ , q

S

= 0 a.e. and |g| = ∞ a.e. Therefore,

⎞1 ⎛ q  ⎝ |g|q dμ⎠ =

1 0

S

⎞1 ⎛

− q1

= ∞.

(i) If S | f | p dμ = 0, (1.4) holds immediately. (ii) If S | f | p dμ > 0, then μ(x : | f (x)| > 0) > 0. Denote A = (x : | f (x)| > 0) ⊆ S, thus   | f g|dμ ≥ | f g|dμ = ∞ · μ(A) = ∞, S

A

which implies (1.4). The desired result is obtained.

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2 Main Results Theorem 2.1 Let f (x) be a nonnegative

and finite measurable function on a measure space (S, B S , μ) with μ(S) < ∞ and S f p (x)dμ(x) < ∞ for all p > 0. Suppose that f 1 (x), f 2 (x), · · · is a sequence of nonnegative measurable functions satisfying conditions 0 ≤ f n (x) ≤ f (x), a.e. μ, x ∈ S, n ≥ 1, and ∞  

f i (x)dμ(x) = ∞.

i=1 S

Then,

sup

⎞ 1 ⎛ 1− p    lim sup ⎝ Tn ( p)dμ(x)⎠ ≤ μ lim sup( f n (x) > 0)

p>0, p=1 n→∞

n→∞

S

⎛ ≤ inf lim inf ⎝



p<0 n→∞

⎞ Tn ( p)dμ(x)⎠

1 1− p

,

S

(2.1) where . ( f n (x) > 0) = (x : f n (x) > 0),

lim sup( f n (x) > 0) = n→∞

∞ ∞  

( f k (x) > 0)

n=1 k=n

and . Tn ( p) = Tn ( p, x) ⎧ p ⎪ n

n

⎨ n f i (x)/ i=1 f i (x)dμ(x) , i f f i (x)dμ(x) > 0, i=1 i=1 = S S ⎪ ⎩ 0, other wise. Here, p is a real number. Proof For an arbitrary positive integer m ≥ 1, let Sm,n =

n  i=m

f i (x), Sn = S1,n =

n 

f i (x), x ∈ S.

i=1

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Then  Sm,n dμ =

n   i=m S

S

 Sn dμ =

f i dμ, S

n  

f i dμ.

i=1 S

Since 0 ≤ f n (x) ≤ f (x), a.e. μ, x ∈ S, n ≥ 1, it follows that Sn = Sm,n + Sm−1 ≤ Sm,n + m f (x), a.e. μ.

(2.2)

For an arbitrary real number p, let . Tm,n ( p) = Tm,n ( p, x)  n p n

n

f i (x)/ i=m f i (x)dμ(x) , i f i=m i=m S f i (x)dμ(x) > 0, S = 0, other wise, and Tm,n = Tm,n (1), Tn = Tn (1). Then, Tn ( p) = T1,n ( p), Tn = T1,n . By (2.2), it follows that Tm,n =

S

Sn − m f (x) Sm,n m f (x)

≥

= Tn −

, a.e. μ, Sn dμ − S Sm−1 dμ S Sn dμ S Sn dμ

and Tm,n ≤

S

Tn Sn



= , a.e. μ, Sn dμ − m S f dμ 1 − m S f dμ/ S Sn dμ

thus m f (x) Tn



Tn −

≤ Tm,n ≤ , a.e. μ. 1 − m S f dμ/ S Sn dμ S Sn dμ Since

∞

i=1 S

f i dμ = ∞, for n large enough

  m S f dμ m f (x) 1−

Tm,n ≤ Tn ≤ Tm,n +

, a.e. μ, S Sn dμ S Sn dμ

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which implies that for any p > 0 



m 1−

f dμ Sn dμ

p 

S S

 

 Tm,n ( p)dμ ≤

S

Tn ( p)dμ ≤ S

S

m f (x) Tm,n +

S Sn dμ

p dμ. (2.3)

By

∞

i=1 S

f i dμ = ∞, we can see that lim



n→∞ S

m f (x) = 0, lim

n→∞ S Sn dμ

Sn dμ = ∞, thus

 p m S f dμ lim 1 −

= 1, m ≥ 1. n→∞ S Sn dμ

For a fixed m,

denote gn (x) = m f (x)/ S Sn dμ. Without loss of generality, we can assume that S f 1 dμ > 0. It is easily seen that 0 ≤ gn (x) ≤ g1 (x) and  S

 m p  p g1 dμ =

f p dμ < ∞. f dμ 1 S S

Hence, for an arbitrary positive number p > 0, by Lemma 1.1, (2.3), it follows that 

 Tm,n ( p)dμ = lim sup

lim sup n→∞ S

S

For all p > 1, q = p/( p − 1) > 1, then 

 Sm,n dμ = S

Tn ( p)dμ, m ≥ 1, p > 0.

n→∞

1 p

+ q1 = 1. By Hölder’s inequality, we have

⎛ ⎞1/ p ⎛ ⎞1/q   p Sm,n I(Sm,n >0) dμ ≤ ⎝ Sm,n dμ⎠ ⎝ I(Sm,n >0) dμ⎠ .

S

S

S

Hence,  μ

n 

 ( f i > 0) =

i=m

 μ

∞ 

i=m

 S

 ( f i > 0)

⎞ 1 q ⎛ 1− p S dμ m,n S I(Sm,n >0) dμ ≥  p 1/ p = ⎝ Tm,n ( p)dμ⎠ , S Sm,n dμ 

S

⎛ ⎛ ⎞ 1 ⎞ 1 1− p 1− p   ≥ lim sup ⎝ Tm,n ( p)dμ⎠ = lim sup⎝ Tn ( p)dμ⎠ , n→∞

n→∞

S

S

⎞ 1 ⎛ 1− p    ∞  ⎠ ⎝ μ lim sup( f n > 0) = lim μ( ( f i > 0)) ≥ lim sup Tn ( p)dμ , n→∞

m→∞

i=m

n→∞

S

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⎞ 1 ⎛ 1− p  ⎠ ⎝ μ lim sup( f n > 0) ≥ sup lim sup Tn ( p)dμ . 



n→∞

p>1 n→∞

S

For all 0 < p < 1, q = p/( p − 1) < 0, then 

 Sm,n dμ = S

1 p

+

1 q

= 1. By Lemma 1.2, we have

⎛ ⎞1/ p ⎛ ⎞1/q   p Sm,n I(Sm,n >0) dμ ≥ ⎝ Sm,n dμ⎠ ⎝ I(Sm,n >0) dμ⎠ .

S

S

S

Hence,  μ

n 







( f i > 0) =

i=m

I(Sm,n >0) dμ ≥ S



S Sm,n dμ 1/ p  p S Sm,n dμ

⎛

q

=⎝



⎞ Tm,n ( p)dμ⎠

1 1− p

,

S

In the same way as proved above, we can obtain  μ

∞ 

 ( f i > 0)

i=m

⎞ 1 ⎞ 1 ⎛ ⎛ 1− p 1− p   ⎠ ⎠ ⎝ ⎝ ≥ lim sup Tm,n ( p)dμ = lim sup Tn ( p)dμ , n→∞

n→∞

S

S

⎞ 1 ⎛ 1− p    ∞  ⎠ ⎝ μ lim sup( f n > 0) = lim μ( ( f i > 0)) ≥ lim sup Tn ( p)dμ , m→∞

n→∞

n→∞

i=m

⎛ ⎞ 1 1− p    μ lim sup( f n > 0) ≥ sup lim sup ⎝ Tn ( p)dμ⎠ . n→∞

0< p<1 n→∞

S

S

Therefore, 



μ lim sup( f n > 0) ≥ n→∞

sup

⎞ 1 ⎛ 1− p  ⎠ ⎝ lim sup Tn ( p)dμ .

p>0, p=1 n→∞

For all p < 0, 0 < q = p/( p − 1) < 1, then can see that 

 Sn dμ = S

123

S

⎛ Sn I(Sn >0) dμ ≥ ⎝

 S

1 p

+

(2.4)

S 1 q

= 1. By Lemma 1.2 again, we

⎞1/ p ⎛ ⎞1/q  ⎝ I(Sn >0) dμ⎠ .

p Sn dμ⎠

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J Theor Probab

Hence,  μ

n 







( f i > 0) =

i=1

I(Sn >0) dμ ≤



S

q



Sn dμ 1/ p p S Sn dμ S

⎛ ⎞ 1 1− p  ⎝ ⎠ = Tn ( p)dμ , S

⎞ 1 ⎛ ∞   n  1− p    ⎠ ⎝ μ ( f i > 0) = lim μ ( f i > 0) ≤ lim inf Tn ( p)dμ , n→∞

i=1





μ lim sup( f n > 0) = lim μ( m→∞

n→∞

n→∞

i=1 ∞ 

( f i > 0)) ≤ μ

∞ 

i=m

S



( f i > 0)

i=1

⎛ ⎞ 1 1− p  ⎝ ⎠ ≤ lim inf Tn ( p)dμ , n→∞

S





μ lim sup( f n > 0) ≤ μ n→∞

∞ 

⎛



( f i > 0) ≤ inf lim inf ⎝



p<0 n→∞

i=1

⎞

1 1− p

Tn ( p)dμ⎠

. (2.5)

S

 

The desired result (2.1) follows from (2.4) and (2.5) immediately. By Theorem 2.1, we can get the following corollaries immediately. Corollary 2.1 Let (S, B S , μ) be a measure space with μ(S) < ∞ and ∞. Then, ⎛ lim sup ⎝

sup



p>0, p=1 n→∞

S



 ≤ μ lim sup An n→∞

⎞

∞

i=1 μ(Ai )

=

1 1− p

Tn ( p)dμ⎠ ⎞ 1 ⎛ 1− p  ⎠ ⎝ ≤ inf lim inf Tn ( p)dμ , p<0 n→∞

S

where Tn ( p) =

 n

i=1 I Ai /

n

i=1 μ(Ai )

0,

p

n , if i=1 μ(Ai ) > 0, other wise.

Here, p is a real number. Proof Take f n = I An for each n ≥ 1 in Theorem 2.1, then ( f n > 0) = An . Corollary 2.1 follows from Theorem 2.1 immediately.  

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Corollary 2.2 Let X be a nonnegative random variable on a probability space (, F, P) with E X p < ∞ for all p > 0. Suppose that X 1 , X 2 , · · · is a sequence of random variables satisfying conditions 0 ≤ X n ≤ X, a.s. P, n ≥ 1, and ∞ 

E X i = ∞.

i=1

Then, sup

lim sup (E Tn ( p))

1 1− p

p>0, p=1 n→∞

  1 ≤ P lim sup(X n > 0) ≤ inf lim inf (E Tn ( p)) 1− p , p<0 n→∞

n→∞

where lim sup(X n > 0) = n→∞

∞ ∞  

(X k > 0),

n=1 k=n

and Tn ( p) =

 n i=1

Xi /

n i=1

E Xi

p

0,

n , if i=1 E X i > 0, other wise.

Here, p is a real number. Remark 2.1 Xie [16] obtained Corollary 2.2 for X ≡ b = const. So our results generalize the corresponding one of Xie [16]. If we take X n = I An and An ∈ F for each n ≥ 1 in Corollary 2.2, then we can get Theorem 1.2 immediately from Corollary 2.2.  Remark 2.2 Let (S, B S , μ) be a measure space with μ(S) < ∞ and ∞ n=1 μ(An ) = ∞. Take f n = I An for each n ≥ 1 in Theorem 2.1, then An = ( f n > 0). When p = 2, one has 

 μ lim sup An n→∞

≥

sup

⎛ ⎛ ⎞ 1 ⎞−1 1− p   lim sup ⎝ Tn ( p)dμ⎠ ≥ lim sup ⎝ Tn (2)dμ⎠

p>0, p=1 n→∞

n→∞

S

n 2 i=1 μ(Ai ) . = lim sup n n→∞ i,k=1 μ(Ai Ak )

123

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J Theor Probab

If we further assume that there exists some real number C1 > 0 such that μ(Ai A j ) ≤ C1 μ(Ai )μ(A j ) for all i = j, then we have 

 μ lim sup An n→∞

≥

1 . C1

(2.6)

Indeed, n



i,k=1 μ(Ai Ak ) 2 i=1 μ(Ai )

Tn (2)dμ =  n S

By the condition

n

2

n n + i=1 μ(Ai ) − C1 i=1 μ2 (Ai ) ≤ n 2 i=1 μ(Ai ) n C1 i=1 μ2 (Ai ) 1 −  = C 1 + n 2 . n i=1 μ(Ai ) i=1 μ(Ai ) C1

i=1 μ(Ai )

∞

n=1 μ(An )

1 → 0, i=1 μ(Ai )

n

= ∞, we can see that

n

2 i=1 μ (Ai ) 2 i=1 μ(Ai )

n

μ(S) → 0 as n → ∞, i=1 μ(Ai )

≤ n

which implies that  lim sup n→∞ S

−1 1 Tn (2)dμ ≥ . C1

Therefore,     1 μ lim sup( f n > 0) = μ lim sup An ≥ C n→∞ n→∞ 1 follows from the inequalities above immediately. , μ) be a measure space with μ(S) < ∞ and An ∈ B S for Theorem 2.2 Let (S, B S  each n ≥ 1. Assume that ∞ n=1 μ(An ) = ∞, and there exist some positive integer n 0 and some real number C2 such that for all N > n ≥ n 0 , N N    μ Ack ≤ C2 μ(Ack ). k=n

k=n

(i) If μ(S) ≤ 1, then   μ lim sup An = μ(S).

(2.7)

n→∞

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(ii) If μ(S) > 1 and lim lim

(μ(S)) N −n+1

n→∞ N →∞

1

e μ(S)

N

k=n

= 0,

μ(Ak )

(2.8)

then (2.7) still holds.   N N  Proof Since μ Ack ≤ C2 μ(Ack ) for all N > n ≥ n 0 , it follows that k=n

k=n

c     = μ(S) − μ lim sup An lim sup An n→∞ n→∞ ∞ ∞     c = μ(S) − lim μ Ak = lim μ Ak

0≤μ

n→∞

≤ lim C2 n→∞

∞ 

n→∞

k=n

k=n

(μ(S) − μ(Ak )) .

k=n

If μ(S) ≤ 1, then lim

∞ 

n→∞

∞ 

(μ(S) − μ(Ak )) ≤ lim

n→∞

k=n

≤ lim

n→∞

k=n ∞ 

(1 − μ(Ak )) e−μ(Ak )

k=n  − ∞ k=n μ(Ak )

= lim e n→∞

= 0.

If μ(S) > 1, then by (2.8), we can get lim

n→∞

∞  k=n

(μ(S) − μ(Ak )) ≤ lim

n→∞

∞ 

μ(S)e−μ(Ak )/μ(S)

k=n

= lim lim

n→∞ N →∞

= lim lim

n→∞ N →∞

N 

μ(S)e−μ(Ak )/μ(S)

k=n

(μ(S)) N −n+1 1

e μ(S)

N

In both cases, we can get (2.7) from the inequalities above.

k=n

μ(Ak )

= 0.  

The following theorem presents a weighted version of the Borel–Cantelli Lemma on the measure space, and the proof of the theorem is inspired by Feng et al. [17].

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Theorem 2.3 Let (S, B S , μ) be a measure space with μ(S) < ∞. If = ∞, where each wn ∈ R, then 

 μ lim sup An n→∞

∞

n=1 wn μ(An )

n 2 k=1 wk μ(Ak ) ≥ lim sup n n . n→∞ i=1 j=1 wi w j μ(Ai A j )

(2.9)

In order to prove (2.9), we need the following lemmas. For any matrix E = (ei j )m×n , denote by (E) the sum of all its entries, i.e. (E) =

m  n 

ei j .

i=1 j=1

Lemma 2.1 Given a partition of an (m + n) × (m + n) symmetric matrix E:  E=

Am×m Cm×n T Cm×n Bn×n

 .

If E is positive semi-definite, then  2 (C) ≤ (A)(B). Proof This lemma follows from the following inequality: ∀ x, y ∈ R, ⎛ ⎞ x ⎜·⎟ ⎜ ⎟ ⎜·⎟ ⎜ ⎟ ⎜·⎟ ⎜ ⎟ ⎜x ⎟ 2 2 ⎟ (x, · · · , x, y, · · · , y)E ⎜ ⎜ y ⎟ = (A)x + 2(C)x y + (B)y ≥ 0, ⎜ ⎟ ⎜·⎟ ⎜ ⎟ ⎜·⎟ ⎜ ⎟ ⎝·⎠ y  

since E is positive semi-definite.

 Lemma 2.2 For each n ≥ 1, the matrix wi w j μ(Ai A j ) n×n is positive semi-definite. Proof It is easy to see that  μ(Ai ) =

 dμ =

Ai

 I (Ai )dμ, μ(Ai A j ) =

S

I (Ai )I (A j )dμ. S

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For each (x1 , x2 , · · · , xn )T ∈ Rn , ⎞ x1 ⎜ x2 ⎟ ⎜ ⎟ n  n  ⎜ · ⎟  ⎟= (x1 , x2 , · · · , xn ) wi w j μ(Ai A j ) ⎜ xi x j wi w j μ(Ai A j ) ⎜ · ⎟ ⎜ ⎟ i=1 j=1 ⎝ · ⎠ xn  n  n  = xi x j wi w j I (Ai )I (A j )dμ ⎛

i=1 j=1

=

=

  n  n S

i=1 j=1

S

i=1

  n

S

xi x j wi w j I (Ai )I (A j )dμ 2

xi wi I (Ai )

dμ ≥ 0.

 

This completes the proof of the lemma. Lemma 2.3 Assume that ∞ 

wn μ(An ) = ∞,

(2.10)

n=1

where each wn ∈ R. Then, for all s ≥ 1, n lim n

n→∞

i=1

i=s+1

n

j=1 wi w j μ(Ai A j )

n

j=s+1 wi w j μ(Ai A j )

= 1.

Proof For each n > s, denote  A = wi w j μ(Ai A j ) 1≤i, j≤s  Bn = wi w j μ(Ai A j ) s+1≤i, j≤n  Cn = wi w j μ(Ai A j ) 1≤i≤s,s+1≤ j≤n    A Cn . E n = wi w j μ(Ai A j ) n×n = CnT Bn Lemma 2.2 has pointed out that E n is positive semi-definite, thus  2 (Cn ) ≤ (A)(Bn )

123

(2.11)

J Theor Probab

follows from Lemma 2.1 immediately. By the Cauchy–Schwarz inequality, we have 

⎛

2

n 

wi μ(Ai )

=⎝

i=s+1

⎡ =⎣



wi I (Ai )dμ⎠

S i=s+1

  n



  n



wi I (Ai ) · I

i=s+1

S

≤

⎞2

n 



=μ

n 

dμ ·

 ⎛ n  Ai · ⎝

i=s+1





i=s+1

S

Ai dμ⎦

i=s+1

2

wi I (Ai )

⎤2



n 

I S

n 

n 

 Ai dμ

i=s+1

⎞

wi w j μ(Ai A j )⎠

i=s+1 j=s+1

≤ μ(S)(Bn ). By (2.10) and the inequality above, we can see that ∞ 

wi μ(Ai ) = ∞,

i=s+1

lim (Bn ) = ∞, for all s ≥ 1,

n→∞

which implies that lim

n→∞

(A) (Cn ) = 0, lim = 0, for all s ≥ 1. n→∞ (Bn ) (Bn )

Thus, for all s ≥ 1, we have n lim n

n→∞

i=1

i=s+1

n

j=1 wi w j μ(Ai A j )

n

j=s+1 wi w j μ(Ai A j )

(A) + (Bn ) + 2(Cn ) n→∞ (Bn )

= lim

= 1 + lim

n→∞

= 1.

(A) (Cn ) + 2 lim n→∞ (Bn ) (Bn )

 

This completes the proof of (2.11). Corollary 2.3 Assume that ∞ 

wn μ(An ) = ∞,

n=1

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J Theor Probab

where each wn ∈ R. Then, for all s ≥ 1, n n 2 2 k=1 wk μ(Ak ) k=s wk μ(Ak ) lim sup n n = lim sup n n . n→∞ n→∞ i=1 j=1 wi w j μ(Ai A j ) i=s j=s wi w j μ(Ai A j ) (2.12) Remark 2.3 According to the proof of Lemma 2.3, we can obtain the following inequality  n 

2



wi μ(Ai )

≤μ

i=s

⎞  ⎛ n n  Ai · ⎝ wi w j μ(Ai A j )⎠ , 1 ≤ s ≤ n

n  i=s

i=s j=s

(2.13) for each wn ∈ R. Furthermore, if {An , n ≥ 1} is a sequence of events defined on a probability space (, F, P), then 

2

n 

wi P(Ai )

≤P

i=s

 ⎛

 n 

Ai

·⎝

i=s

n  n 

⎞ wi w j P(Ai A j )⎠ , 1 ≤ s ≤ n.

i=s j=s

(2.14) In particular,  n 

2 P(Ai )

 ≤P

i=1

n 

 ⎛ Ai

·⎝

i=1

n  n 

⎞ P(Ai A j )⎠ , n ≥ 1.

(2.15)

i=1 j=1

The above inequality (2.15) is the classical Chung–Erdös inequality (see Chung and Erdös [18]). So the inequalities (2.13) and (2.14) can be viewed as weighted version of the Chung–Erdös type inequality. Proof of Theorem 2.3 By (2.13) and (2.12), we can get 

 μ lim sup An n→∞

=μ

∞ ∞ 





lim μ

= lim

s→∞

n→∞



= lim μ

Ak

s=1 k=s

s→∞



n  k=s



∞ 

 Ak

k=s

Ak

 n 2 k=s wk μ(Ak ) ≥ lim lim sup n n s→∞ n→∞ i=s j=s wi w j μ(Ai A j )   n 2 k=1 wk μ(Ak ) = lim lim sup n n s→∞ n→∞ i=1 j=1 wi w j μ(Ai A j )

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J Theor Probab

n 2 k=1 wk μ(Ak ) . = lim sup n n n→∞ i=1 j=1 wi w j μ(Ai A j )  

This completes the proof of the theorem. By Theorem 2.3, we can get the following corollaries immediately.

Corollary 2.4 Let (S, B S , μ) be a measure space with μ(S) < ∞ and μ(An ) > 0 for all n ≥ 1. Then, 



≥ lim sup 

μ lim sup An n→∞

n i=1

n→∞

n

n2

μ(Ai A j ) j=1 μ(Ai )μ(A j )

.

Proof Take wn = 1/μ(An ) for all n ≥ 1, then the desired result follows from Theorem 2.3 immediately.   Corollary 2.5 Let {An , n ≥ 1} be a sequence of events defined on a probability space (, F, P). Assume that ∞ 

wn P(An ) = ∞,

n=1

where each wn ∈ R. Then, 

 P lim sup An n→∞

n 2 k=1 wk P(Ak ) ≥ lim sup n n . n→∞ i=1 j=1 wi w j P(Ai A j )

The following corollary follows from Corollary 2.4 or Corollary 2.5 immediately. Corollary 2.6 Let {An , n ≥ 1} be a sequence of events defined on a probability space (, F, P). Assume that P(An ) > 0 holds for all n ≥ 1. Then, 

 P lim sup An n→∞

≥ lim sup  n→∞

n i=1

n

n2

P(Ai A j ) j=1 P(Ai )P(A j )

.

3 Example Example 3.1 Let (S,  B S , μ) be a measure space with μ(S) < ∞, μ(A)μ(B) > . μ(AB) = 0 and μ(A B) = μ(S). For an arbitrary positive integer n, A3n−2 = A, . A3n−1 = A3n = B. Denote f i = I Ai for each i ≥ 1. Then,

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J Theor Probab

⎞ 1 ⎛ 1− p    ⎠ ⎝ lim sup Tn ( p)dμ = μ lim sup An

sup

p>0, p=1 n→∞

n→∞

S

⎛ = inf lim inf ⎝



p<0 n→∞

⎞

1 1− p

Tn ( p)dμ⎠

= μ(S),

S

(3.1) where Tn ( p) is defined in Theorem 2.1. Proof For p > 0, since μ(AB) = 0, we have 

p S3n dμ

S

=

  3n

 dμ = n p

I Ai

i=1

S

=n

p

(I A + 2I B ) p dμ S−AB





p



I A + 2 I B dμ = n p

 p

S−AB



I A + 2 p I B dμ

S

3n    p p = n μ(A) + 2 μ(B) , f i dμ = n (μ(A) + 2μ(B)) . i=1 S

Thus,

 T3n ( p)dμ = S

p S S3n dμ 

p 3n i=1 S f i dμ

=

μ(A) + 2 p μ(B) , p > 0. (μ(A) + 2μ(B)) p

(3.2)

   Since μ(AB) = 0 and μ (A B)c = μ(S) − μ(A B) = 0, we have for p < 0 



p

S3n dμ = n p S

(I A + 2I B ) p dμ S

=n



 (I A + 2I B ) dμ + n

p

p

A





(I A + 2I B ) p dμ 

(A  = np A





= np S

123

AB

B−AB

+n p

(I A + 2I B ) p dμ

p

B)

c



I A + 2 p I B dμ

B−AB



 I A + 2 p I B dμ = n p μ(A) + 2 p μ(B) ,

J Theor Probab

which implies that



p

S dμ μ(A) + 2 p μ(B) T3n ( p)dμ =  S 3n , p < 0. p = 3n (μ(A) + 2μ(B)) p f dμ i=1 S i

S

(3.3)

By (3.2) and (3.3), we can get ⎛ ⎝



⎞ T3n ( p)dμ⎠

S

1 1− p

#

μ(A) + 2 p μ(B) = (μ(A) + 2μ(B)) p

$

1 1− p

, p = 1.

(3.4)

Obviously, the right-hand side of (3.4) is continuous at p = 0, hence ⎛ sup

lim sup ⎝

p>0, p=1 n→∞



⎞

1 1− p

Tn ( p)dμ⎠

S

#

μ(A) + 2 p μ(B) ≥ lim p↓0 (μ(A) + 2μ(B)) p

$

1 1− p

= μ(A) + μ(B) = μ(S),

(3.5)

and ⎛ inf lim inf ⎝



p<0 n→∞

⎞

1 1− p

Tn ( p)dμ⎠

S

#

μ(A) + 2 p μ(B) ≤ lim p↑0 (μ(A) + 2μ(B)) p

$

1 1− p

= μ(A) + μ(B) = μ(S). Therefore, (3.1) follows from (3.5), (3.6) and Theorem 2.1 immediately.

(3.6)  

Example 3.2 Let (S, B S , μ) be a measure space with μ(S) < ∞ and μ(A)μ(B) > 0. For an arbitrary positive integer n, denote A3n−2 = A, A3n−1 = B, A3n = AB. Then, 

 μ lim sup An n→∞

= μ(A



B).

Applying Theorem 2.3 with the following weight sequence 1, 1, −1, 1, 1, −1, 1, 1, −1, · · · , we can get    B). μ lim sup An ≥ μ(A) + μ(B) − μ(AB) = μ(A n→∞

123

J Theor Probab

   In fact, μ lim sup An = μ(A B). Thus, Theorem 2.3 gives the best estimation n→∞

for this example. Acknowledgments The authors are most grateful to the Editor-in-Chief James Allen Fill and anonymous associate Editor and referee for careful reading of the manuscript and valuable suggestions which helped significantly improving an earlier version of this paper.

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