定义1.1  称随机变量$ \{X_{i};1\leq i\leq n\} $是NA (negatively associated)的, 若对集合$ \{1, 2, \cdots, n\} $中的任意不相交的两个非空子集$ A_{1} $, $ A_{2} $, 都有

$ \begin{eqnarray*} {\rm Cov}\big(f_{1}(X_{i}, i\in A_{1}), f_{2}(X_{j}, j\in A_{2})\big)\leq 0, \end{eqnarray*} $

其中$ f_{1} $和$ f_{2} $是使得上式成立且对每个变元都为非降（或非升）的函数.称随机变量序列$ \{X_{n}; n\geq 1\} $是NA序列, 若对任意的$ n\geq2 $, $ \{X_{i};1\leq i\leq n\} $都是NA的.

定义1.2  称随机变量序列$ \{X_{n};n \geq 1\} $是AANA (asymptotically almost negatively associated)的, 若存在非负序列$ \mu(n)\rightarrow 0 $, $ n\rightarrow \infty $, 对任意的$ n, k\geq 1 $, 都有

$ \begin{eqnarray*} {\rm Cov}\big(f_{1}\big(X_{n}\big), f_{2}\big(X_{n+1}, \cdots, X_{n+k}\big)\big) \leq \mu(n)\big({\rm Var}\big(f_{1}\big(X_{n}\big)\big){\rm Var}\big(f_{2}\big(X_{n+1}, \cdots , X_{n+k}\big)\big)\big)^{1/2}, \end{eqnarray*} $

其中$ f_{1} $, $ f_{2} $为任何两个使该协方差存在且对每个变元都是单调非降的函数.称$ \{\mu(n); n \geq 1\} $为该AANA序列的混合系数.

Chandra和Ghosal^[2]首先提出了AANA序列的概念, 可以明显看出, AANA序列真包含NA序列($ \mu(n) = 0, n\geq 1 $). Joag-Dev和Proschan^[7]指出并证明, 许多多元分布具有NA性质.因此, 将NA随机变量的极限性质推广到AANA随机变量的情形具有重要意义.关于AANA随机变量的极限性质的更多细节, 参见文献[6, 8, 12, 15, 17-18]等等.

Hsu和Robbins^[5]首先介绍了完全收敛性的概念.称随机序列$ \{X_{n};n\geq 1\} $完全收敛于常数$ {\lambda} $, 如果对任意的$ {\varepsilon>0} $, 有$ \sum\limits_{n = 1}^\infty P(|X_{n}-\lambda|>\varepsilon)<\infty $.随机变量的完全收敛性是建立几乎处处收敛的重要工具, 关于它的第一个结果是由Hsu和Robbins^[5]以及Erdös^[4]给出的.许多作者把这个基本定理推广和改进到了多个方向, 如文献[9-11, 13-14, 16]等等.特别地, Baek等^[1]研究了NA随机变量加权和的完全收敛性, 得到如下结果.

定理1.1  设$ \left\{X_{ni}; i\geq 1, n\geq 1\right\} $为行间NA随机变量阵列, 且对任意的$ i\geq 1, n\geq 1 $和$ x\geq 0 $, 有$ EX_{ni} = 0 $, $ P\big(\big|X_{ni}\big|\geq x\big)\leq CP\big(|X|\geq x\big) $.又设$ \beta\geq -1 $, $ \left\{a_{ni}; i\geq 1, n\geq 1\right\} $为常数阵列, 满足:$ { } \sup_{i\geq1}\big|a_{ni}\big| = O\big(n^{-r}\big) $, $ r>0 $和$ \sum\limits_{i = 1}^\infty \big|a_{ni}\big| = O\big(n^{\alpha}\big) $, $ \alpha\in [0, r) $.

$ \big({\rm i}\big) $当$ 1+\alpha+\beta>0 $时, 假设存在$ \delta>0 $使得$ 1+\alpha/r<\delta\leq 2 $.令

$ s = \max \{1+(1+\alpha+\beta)/r, \delta\}, $

若$ E|X|^{s}<\infty $, 则对任意的$ \varepsilon>0 $,

(1.1) $ \begin{equation} \sum\limits_{n = 1}^\infty n^{\beta}P\bigg(\bigg|\sum\limits_{i = 1}^{\infty}a_{ni}X_{ni}\bigg|>\varepsilon\bigg)<\infty. \end{equation} $

$ \big({\rm ii}\big) $当$ 1+\alpha+\beta = 0 $时, 若$ E\big(|X|{\rm log}\big(1+|X|\big)\big)<\infty $, 则(1.1)式成立.

Wang等^[12]把上述结果从行间NA随机变量推广到了AANA随机变量, 而且增加考虑了$ 1+\alpha+\beta<0 $的情况, 获得如下结果.

定理1.2  设$ \beta\geq -1 $, $ \left\{X_{ni}; i\geq 1, n\geq 1\right\} $为行间AANA随机变量阵列, 且被随机变量$ X $随机控制.假设常数阵列$ \left\{a_{ni}; i\geq 1, n\geq 1\right\} $满足如下条件:

(1.2) $ \begin{equation} \sup\limits_{i\geq1}\big|a_{ni}\big| = O\big(n^{-r}\big), \; r>0, \end{equation} $

(1.3) $ \begin{equation} \sum\limits_{i = 1}^{\infty}\big|a_{ni}\big|^{\theta} = O\big(n^{\alpha}\big), \;\mbox{其中}\ 0<\theta<2, \; \theta+\frac{\alpha}{r}<2. \end{equation} $

$ \big({\rm i}\big) $假设当$ 1<\theta<2 $时, $ \sum\limits_{n = 1}^{\infty}\mu^{2}(n)<\infty $.若$ 1+\alpha+\beta<0 $, $ E|X|^{\theta}<\infty $, 则对任意的$ \varepsilon>0 $,

(1.4) $ \begin{equation} \sum\limits_{n = 1}^\infty n^{\beta}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>\varepsilon\bigg)<\infty, \end{equation} $

且(1.1)式也成立.

$ \big({\rm ii}\big) $若$ 1+\alpha+\beta = 0 $, 且

(1.5) $ \begin{equation} E|X|^{\theta}{\rm log}|X|<\infty, \end{equation} $

进一步假设$ EX_{ni} = 0 $, 当$ 1\leq\theta<2 $时, $ \sum\limits_{n = 1}^{\infty}\mu^{2}(n)<\infty $, 则(1.1)式和(1.4)式成立.

$ \big({\rm iii}\big) $若$ 1+\alpha+\beta>0 $, 设$ \beta>-1 $, 且当$ s = \theta+(1+\alpha+\beta)/r $时, $ E|X|^{s}<\infty $.进一步假设$ EX_{ni} = 0 $, $ \sum\limits_{n = 1}^{\infty}\mu^{1/(p-1)}(n)<\infty $, $ p\in \big(3\cdot2^{k-1}, 4\cdot2^{k-1}\big] $, 并且

(1.6) $ \begin{equation} p>\max\left\{2, s, \frac{2\big(1+\beta\big)}{r\big(2-\theta\big)-\alpha}\right\}, \end{equation} $

其中$ s\geq1 $, 正整数$ k\geq1 $, 则(1.1)式和(1.4)式成立.

该文在Baek等^[1]和Wang等^[12]研究结果的基础之上, 将进一步研究不同分布条件下行间AANA随机变量加权和的完全矩收敛.所得结果推广和改进了Baek等^[1]和Wang等^[12]的结论.

定义1.3  称随机变量序列$ \left\{Z_{n}; n\geq 1\right\} $完全矩收敛, 若对$ a_{n}>0, b_{n}>0, q>0 $,

$ \sum\limits_{n = 1}^{\infty}a_{n}E\big(b_{n}^{-1}\big|Z_{n}\big|-\varepsilon\big)_{+}^{q}<\infty, \;\varepsilon>0. $

完全矩收敛的概念是由Chow^[3]首先提出的.

定义1.4  称随机变量序列$ \left\{X_{n}; n\geq 1\right\} $被随机变量$ X $随机控制, 若存在正常数$ C $, 使得对所有的$ n\geq 1 $, $ x\geq 0 $, 有$ P\big(\big|X_{n}\big|\geq x\big)\leq CP\big(|X|\geq x\big). $

引理2.1^[17]  设$ \left\{X_{n}; n\geq 1\right\} $是混合系数为$ \left\{\mu (n); n\geq 1\right\} $的AANA随机变量序列, 设$ \left\{f_{n}; n\geq 1\right\} $是单调不增（或单调不减）的函数序列, 则$ \left\{f_{n}\big(X_{n}\big); n\geq 1\right\} $仍是混合系数为$ \left\{\mu (n); n\geq 1\right\} $的AANA随机变量序列.

引理2.2^[17]  设$ \left\{X_{n}; n\geq 1\right\} $是一均值为0的AANA随机变量序列, 当$ 1<p\leq2 $时, 混合系数$ \left\{\mu (n); n\geq 1\right\} $满足$ \sum\limits_{n = 1}^{\infty}\mu^{2}(n)<\infty $, 则存在只依赖于$ p $的正常数$ C = C_{p} $, 使得

(2.1) $ \begin{equation} E\bigg(\max\limits_{1\leq k\leq n}\bigg|\sum\limits_{i = 1}^k X_{i}\bigg|^p \bigg)\leq C\sum\limits_{i = 1}^nE\big|X_{i}\big|^{p}. \end{equation} $

当$ p\in \big(3\cdot2^{k-1}, 4\cdot2^{k-1}\big] $时, $ k $是正整数, $ \sum\limits_{n = 1}^{\infty}\mu^{1/(p-1)}(n)<\infty $, 则存在只依赖于$ p $的正常数$ C = C_{p} $, 使得

(2.2) $ \begin{equation} E\bigg(\max\limits_{1\leq k\leq n}\bigg|\sum\limits_{i = 1}^k X_{i}\bigg|^p \bigg)\leq C \left\{\sum\limits_{i = 1}^n E\big|X_{i}\big|^p+\bigg(\sum\limits_{i = 1}^n EX_{i}^2\bigg)^{p/2} \right\}. \end{equation} $

引理2.3^[16]  设$ \{X_{n};n\geq 1\} $为一随机变量序列, 且被随机变量$ X $随机控制, 则对所有的$ u>0 $, $ t>0 $, 下面两式成立:

(2.3) $ \begin{equation} E\big|X_{n}\big|^u I\big(\big|X_{n}\big|\leq t\big)\leq C\left[E|X|^u I\big(|X|\leq t\big)+t^u P\big(|X|>t\big)\right], \end{equation} $

(2.4) $ \begin{equation} E\big|X_{n}\big|^u I\big(\big|X_{n}\big|> t\big)\leq CE|X|^u I\big(|X|> t\big). \end{equation} $

整篇文章中, $ \left\{X_{ni};i\geq1, n\geq1\right\} $为行间AANA阵列, 且混合系数为$ \left\{\mu(i), i\geq1\right\} $.正常数$ C $在不同的地方有所不同.

定理3.1  设$ \beta\geq -1 $, $ \left\{X_{ni};i\geq1, n\geq1\right\} $为行间AANA阵列, 且被随机变量$ X $随机控制.$ \left\{a_{ni};i\geq1, n\geq1\right\} $为满足(1.2)式, (1.3)式的常数阵列.

$ \big({\rm i}\big) $设$ 1<\theta<2 $, $ \sum\limits_{n = 1}^{\infty}\mu^{2}(n)<\infty $.若$ 1+\alpha+\beta<0 $, $ E|X|^{\theta}<\infty $, 则对任意的$ \varepsilon>0 $,

(3.1) $ \begin{equation} \sum\limits_{n = 1}^\infty n^{\beta}E\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|-\varepsilon\bigg)_{+}^{\theta}<\infty, \end{equation} $

以及

(3.2) $ \begin{equation} \sum\limits_{n = 1}^\infty n^{\beta}E\bigg(\bigg|\sum\limits_{i = 1}^{\infty}a_{ni}X_{ni}\bigg|-\varepsilon\bigg)_{+}^{\theta}<\infty. \end{equation} $

$ \big({\rm ii}\big) $若$ 1+\alpha+\beta = 0 $, 且

(3.3) $ \begin{equation} E|X|^{\theta}{\rm log}|X|<\infty, \end{equation} $

进一步假设$ EX_{ni} = 0 $, 当$ 1\leq\theta<2 $时, $ \sum\limits_{n = 1}^{\infty}\mu^{2}(n)<\infty $, 则(3.1)式和(3.2)式成立.

$ \big({\rm iii}\big) $若$ 1+\alpha+\beta>0 $, $ \beta>-1 $以及

(3.4) $ \begin{equation} E|X|^{s}<\infty, \; s = \theta+\frac{1+\alpha+\beta}{r}. \end{equation} $

进一步假设$ EX_{ni} = 0 $, $ \sum\limits_{n = 1}^{\infty}\mu^{1/(p-1)}(n)<\infty $, $ p\in (3\cdot2^{k-1}, 4\cdot2^{k-1}] $, 并且

(3.5) $ \begin{equation} p>\max\left\{2, s, \frac{2\big(1+\beta\big)}{r\big(2-\theta\big)-\alpha}\right\}, \end{equation} $

其中$ s\geq1 $, $ k $为正整数, 则(3.1)式和(3.2)式成立.

注3.1  定理3.1不仅把定理1.1关于行间NA阵列的结果扩展到了AANA阵列的情况, 而且考虑了$ 1+\alpha+\beta<0 $的情形, 从而补充了Baek等^[1]的结论.

注3.2  在定理3.1的条件下, 对任意$ \varepsilon>0 $,

(3.6) $ \begin{eqnarray} \infty &>&\sum\limits_{n = 1}^\infty n^{\beta}E\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|-\varepsilon\bigg)_{+}^{\theta}\\ & = &\sum\limits_{n = 1}^\infty n^{\beta}\int_{0}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|-\varepsilon>x^{1/ \theta}\bigg){\rm d}x\\ &\geq&C\int_{0}^{1}\sum\limits_{n = 1}^\infty n^{\beta}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>\varepsilon\bigg){\rm d}x\\ & = &C\sum\limits_{n = 1}^\infty n^{\beta}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>\varepsilon\bigg). \end{eqnarray} $

从(3.6)式可以看出, 完全矩收敛蕴含完全收敛.比较Wang等^[12]的结果, 在同样的条件下, 定理3.1的结果比定理1.2的结果更强.因此, 定理3.1提高了Wang等^[12]的结论.

证  (3.2)式的证明完全类似于(3.1)式的证明, 因此, 该文仅证明(3.1)式成立.不失一般性, 设$ a_{ni}>0 $, $ n, i\geq1 $ (否则, 用$ a_{ni}^{+} $和$ a_{ni}^{-} $替代$ a_{ni} $, 则$ a_{ni} = a_{ni}^{+}-a_{ni}^{-} $).由(1.2)式和(1.3)式, 不妨设$ { } \sup_{i\geq1}\big|a_{ni}\big|\leq n^{-r} $和$ \sum\limits_{i = 1}^{\infty}\big|a_{ni}\big|^{\theta}\leq n^{\alpha} $.对任意$ \varepsilon>0 $,

(3.7) $ \begin{eqnarray} && \sum\limits_{n = 1}^\infty n^{\beta}E\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|-\varepsilon\bigg)_{+}^{\theta} = \sum\limits_{n = 1}^{\infty} n^{\beta}\int_{0}^{1}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>\varepsilon+x^{1/ \theta}\bigg){\rm d}x \\ &&+\sum\limits_{n = 1}^{\infty} n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>\varepsilon+x^{1/ \theta}\bigg){\rm d}x \\ &\leq&\sum\limits_{n = 1}^\infty n^{\beta}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>\varepsilon\bigg) +\sum\limits_{n = 1}^{\infty} n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>x^{1/ \theta}\bigg){\rm d}x \\ &\triangleq& I+J. \end{eqnarray} $

根据Wang等^[12]中的定理3.1的证明, 可以直接得到$ I<\infty $.因此, 只需分如下三种情况证明$ J<\infty $.

$ \big({\rm i}\big) $$ 1+\alpha+\beta<0 $

注意到:

(3.8) $ \begin{equation} \int_{a}^{\infty}P\big(\big|Y\big|>x^{1/ \theta}\big){\rm d}x\leq E\big|Y\big|^{\theta}I\big(\big|Y\big| >a^{1/ \theta}\big). \end{equation} $

由(3.8)式, (2.1)式和(1.3)式可知

(3.9) $ \begin{eqnarray} J &\leq& \sum\limits_{n = 1}^\infty n^{\beta}E\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|^{\theta}\bigg) I\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|^{\theta}>1\bigg)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}E\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|^{\theta}\bigg)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\big|a_{ni}\big|^{\theta}E|X|^{\theta} \leq C\sum\limits_{n = 1}^\infty n^{\alpha+\beta}E|X|^{\theta} \leq CE|X|^{\theta}<\infty. \end{eqnarray} $

接下来证明当$ 1+\alpha+\beta\geq0 $时, $ J<\infty $.对固定的$ n\geq1 $, 定义

$ Y_{ni} = -x^{1/ \theta}I\big(a_{ni}X_{ni}<-x^{1/\theta}\big)+a_{ni}X_{ni}I\big(\big|a_{ni}X_{ni}\big|\leq x^{1/\theta}\big)+x^{1/ \theta}I\big(a_{ni}X_{ni}>x^{1/\theta}\big), $

$ A = \bigcap\limits_{i = 1}^{n}\big(Y_{ni} = a_{ni}X_{ni}\big), \; \; B = \overline {A} = \bigcup\limits_{i = 1}^{n}\big(Y_{ni}\neq a_{ni}X_{ni}\big) = \bigcup\limits_{i = 1}^{n}\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big), $

$ E_{n} = \bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>x^{1/ \theta}\bigg). $

由引理2.1可知, $ \left\{Y_{ni}; i\geq1, n\geq 1\right\} $仍是AANA随机阵列, 容易验证

$ E_{n} = E_{n}A\cup E_{n}B\subset \bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Y_{ni}\bigg|>x^{1/ \theta}\bigg)\cup \bigg(\bigcup\limits_{i = 1}^{n}\big(\big|a_{ni}X_{ni}\big|>x^{1/ \theta}\big)\bigg), $

则

(3.10) $ \begin{eqnarray} J & = &\sum\limits_{n = 1}^{\infty} n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}a_{ni}X_{ni}\bigg|>x^{1/ \theta}\bigg){\rm d}x\\ &\leq& \sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}P\big(\big|a_{ni}X_{ni}\big|>x^{1/ \theta}\big){\rm d}x +\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Y_{ni}\bigg|>x^{1/ \theta}\bigg){\rm d}x\\ &\triangleq& J_{1}+J_{2}. \end{eqnarray} $

因此, 要证(3.1)式成立, 只需证明$ J_{1}<\infty $, $ J_{2}<\infty $.

$ \big({\rm ii}\big) $$ 1+\alpha+\beta = 0 $

根据(1.2)式, 不失一般性, 不妨设$ \big|a_{ni}\big|\leq n^{-r} $.因此, 由(3.8), (1.2), (1.3)和(3.3)式可知

(3.11) $ \begin{eqnarray} J_{1} &\leq& \sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n E\big|a_{ni}X_{ni}\big|^{\theta}I\big(\big|a_{ni}X_{ni}\big|>1\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\big|a_{ni}\big|^{\theta}E|X|^{\theta} I\big(|X|>\big|a_{ni}\big|^{-1}\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\alpha+\beta}E|X|^{\theta}I\big(|X|>n^{r}\big)\\ & = & C\sum\limits_{n = 1}^\infty n^{-1}\sum\limits_{k = n}^\infty E|X|^{\theta} I\big(k^{r}<|X|\leq\big(k+1\big)^{r}\big)\\ &\leq& C\sum\limits_{k = 1}^\infty E|X|^{\theta}I\big(k^{r}<|X|\leq\big(k+1\big)^{r}\big) \sum\limits_{n = 1}^kn^{-1}\\ &\leq& C\sum\limits_{k = 1}^\infty {\rm log}{k}E|X|^{\theta}I\big(k^{r}<|X|\leq\big(k+1\big)^{r}\big)\\ &\leq& CE|X|^{\theta}{\rm log}{|X|}<\infty. \end{eqnarray} $

下面分$ 0<\theta<1 $和$ 1\leq \theta<2 $两种情况来证明$ J_{2}<\infty $.

情况1  $ 0<\theta<1 $

取某个$ p $使得$ 1<p\leq2 $, 根据Markov不等式, $ C_{r} $不等式以及引理2.3可得

(3.12) $ \begin{eqnarray} J_{2} & = &\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Y_{ni}\bigg| >x^{1/ \theta}\bigg){\rm d}x\\ &\leq& \sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-p/ \theta}E\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Y_{ni}\bigg|^{p}\bigg){\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/ \theta}E\big|Y_{ni}\big|^{p}{\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/ \theta}\left[E\big|a_{ni}X_{ni}\big|^{p} I\big(\big|a_{ni}X_{ni}\big|\leq x^{1/\theta}\big) +x^{p/\theta}P\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\right]{\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/\theta}E\big|a_{ni}X\big|^{p} I\big(\big|a_{ni}X\big|\leq 1\big){\rm d}x\\ &&+ C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/\theta}E\big|a_{ni}X\big|^{p} I\big(1<\big|a_{ni}X\big|\leq x^{1/\theta}\big){\rm d}x\\ &&+ C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}P \big(\big|a_{ni}X\big|>x^{1/\theta}\big){\rm d}x\\ &\triangleq& J_{21}+J_{22}+J_{23}. \end{eqnarray} $

首先证明$ J_{22}<\infty $, $ J_{23}<\infty $.令$ x = t^{\theta} $, 则

(3.13) $ \begin{eqnarray} J_{22} & = & C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/\theta}E\big|a_{ni}X\big|^{p} I\big(1<\big|a_{ni}X\big|\leq x^{1/\theta}\big){\rm d}x\\ & = & C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}t^{\theta-p-1}E\big|a_{ni}X\big|^{p} I\big(1<\big|a_{ni}X\big|\leq t\big){\rm d}t\\ & = & C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\sum\limits_{k = 1}^\infty\int_{k}^{k+1}t^{\theta-p-1}E\big|a_{ni}X\big|^{p} I\big(1<\big|a_{ni}X\big|\leq t\big){\rm d}t\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\sum\limits_{k = 1}^\infty k^{\theta-p-1}E\big|a_{ni}X\big|^{p} I\big(1<\big|a_{ni}X\big|\leq k+1\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\sum\limits_{k = 1}^\infty k^{\theta-p-1}\sum\limits_{m = 1}^kE\big|a_{ni}X\big|^{p} I\big(m<\big|a_{ni}X\big|\leq m+1\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\sum\limits_{m = 1}^\infty E\big|a_{ni}X\big|^{p} I\big(m<\big|a_{ni}X\big|\leq m+1\big)\sum\limits_{k = m}^\infty k^{\theta-p-1}\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\sum\limits_{m = 1}^\infty m^{\theta-p}E\big|a_{ni}X\big|^{p} I\big(m<\big|a_{ni}X\big|\leq m+1\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{\theta}I\big(\big|a_{ni}X\big|>1\big)<\infty. \end{eqnarray} $

由(3.8)式以及$ J_{1}<\infty $的证明过程, 易知

(3.14) $ \begin{equation} J_{23} \leq C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{\theta}I\big(\big|a_{ni}X\big|>1\big)<\infty. \end{equation} $

其次证明$ J_{21}<\infty $.

(3.15) $ \begin{eqnarray} J_{21} & = & C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/\theta}E\big|a_{ni}X\big|^{p} I\big(\big|a_{ni}X\big|\leq 1\big){\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{p}I\big(\big|a_{ni}X\big|\leq1\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{p}I\big(|X|\leq\big|a_{ni}\big|^{-1}\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{p}I\big(|X|\leq n^{r}\big) +C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{p}I\big(n^{r}<|X|\leq \big|a_{ni}\big|^{-1}\big)\\ &\triangleq& J_{21}^{*}+J_{21}^{**}. \end{eqnarray} $

根据$ \theta<p $, $ r>0 $, (1.2)和(1.3)式可知

(3.16) $ \begin{eqnarray} J_{21}^{*} &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\big|a_{ni}\big|^{\theta}\big(\sup\limits_{i\geq1}\big|a_{ni}\big|\big)^{p-\theta} E|X|^{p}I\big(|X|\leq n^{r}\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\alpha+\beta-r(p-\theta)}E|X|^{p}I\big(|X|\leq n^{r}\big)\\ & = & C\sum\limits_{n = 1}^\infty n^{-1-r(p-\theta)}\sum\limits_{k = 1}^nE|X|^{p}I\big((k-1)^{r}<|X|\leq k^{r}\big)\\ & = & C\sum\limits_{k = 1}^\infty E|X|^{p}I\big((k-1)^{r}<|X|\leq k^{r}\big)\sum\limits_{n = k}^\infty n^{-1-r(p-\theta)}\\ &\leq& C\sum\limits_{k = 1}^\infty k^{-r(p-\theta)}E|X|^{p}I\big((k-1)^{r}<|X|\leq k^{r}\big)\\ &\leq& CE|X|^{p+\frac{-r(p-\theta)}{r}} = CE|X|^{\theta}<\infty. \end{eqnarray} $

由$ J_{1}<\infty $的证明过程, 再次可得

(3.17) $ \begin{eqnarray} J_{21}^{**} &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{\theta}I\big(n^{r}<|X|\leq \big|a_{ni}\big|^{-1}\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{\theta}I\big(|X|\geq n^{r}\big)<\infty. \end{eqnarray} $

情况2  $ 1\leq\theta<2 $

首先证明

(3.18) $ \begin{equation} \sup\limits_{x\geq 1}x^{-1/\theta}\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}EY_{ni}\bigg|\rightarrow 0, \; n\rightarrow \infty. \end{equation} $

事实上, 由条件$ EX_{ni} = 0 $, $ Y_{ni} $的定义, 引理2.3, (1.2)和(1.3)式以及$ E|X|^{\theta}<\infty $可得

(3.19) $ \begin{eqnarray} \sup\limits_{x\geq 1}x^{-1/\theta}\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}EY_{ni}\bigg| &\leq& C\sup\limits_{x\geq 1}x^{-1/\theta}\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Ea_{ni}X_{ni}I\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\bigg|\\ &&+ C\sup\limits_{x\geq 1}x^{-1/\theta}\sum\limits_{i = 1}^{n}x^{1/\theta}P\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\\ &\leq& C\sup\limits_{x\geq 1}x^{-1/\theta}\sum\limits_{i = 1}^{n}E\big|a_{ni}X\big|I\big(\big|a_{ni}X\big|>x^{1/\theta}\big)\\ &\leq& C\sup\limits_{x\geq 1}x^{-1}\sum\limits_{i = 1}^{n}E\big|a_{ni}X\big|^{\theta}I\big(\big|a_{ni}X\big|>x^{1/\theta}\big)\\ &\leq& C\sum\limits_{i = 1}^{n}E\big|a_{ni}X\big|^{\theta}I\big(\big|a_{ni}X\big|>1\big)\\ &\leq& Cn^{-1-\beta}E|X|^{\theta}I\big(|X|>n^{r}\big)\rightarrow 0, \; n\rightarrow \infty. \end{eqnarray} $

因此, 对足够大的$ n $, 有

(3.20) $ \begin{equation} \max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}EY_{ni}\bigg|\leq \frac{x^{1/\theta}}{2}. \end{equation} $

根据(3.20)式, 引理2.2, Markov不等式以及情况1中$ J_{2}<\infty $ (用2替代指数$ p $)的证明过程, 可得

(3.21) $ \begin{eqnarray} J_{2} & = &\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Y_{ni}\bigg| >x^{1/ \theta}\bigg){\rm d}x\\ &\leq& \sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}P \bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}\big(Y_{ni}-EY_{ni}\big)\bigg|>\frac {x^{1/ \theta}}{2}\bigg){\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-2/ \theta}E \bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}\big(Y_{ni}-EY_{ni}\big)\bigg|^{2}\bigg){\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-2/ \theta}EY_{ni}^2{\rm d}x<\infty. \end{eqnarray} $

$ \big({\rm iii}\big) $$ 1+\alpha+\beta>0 $

由(3.8)式, 引理2.3, (1.2)式, (1.3)式和(3.4)式, 可知

(3.22) $ \begin{eqnarray} J_{1} & = &\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}P\big(\big|a_{ni}X_{ni}\big|>x^{1/ \theta}\big){\rm d}x\\ &\leq& \sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{\theta}I\big(\big|a_{ni}X\big|>1\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\alpha+\beta}E|X|^{\theta}I\big(|X|>n^{r}\big)\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\alpha+\beta}\sum\limits_{k = n}^\infty E|X|^{\theta}I\big(k^{r}<|X|\leq \big(k+1\big)^{r}\big)\\ &\leq& C\sum\limits_{k = 1}^\infty E|X|^{\theta}I\big(k^{r}<|X|\leq \big(k+1\big)^{r}\big)\sum\limits_{n = 1}^k n^{\alpha+\beta}\\ &\leq& C\sum\limits_{k = 1}^\infty k^{1+\alpha+\beta}E|X|^{\theta}I\big(k^{r}<|X|\leq \big(k+1\big)^{r}\big)\\ &\leq& CE|X|^{\theta+(1+\alpha+\beta)/r}<\infty. \end{eqnarray} $

下面分$ s\geq 1 $和$ s<1 $两种情况证明$ J_{2}<\infty $.

情况1  $ s\geq 1 $

首先证明(3.18)式仍成立.根据$ Y_{ni} $的定义, $ EX_{ni} = 0 $, 引理2.3, (1.2)式, (1.3)式以及$ E|X|^{s}<\infty $, 可得

(3.23) $ \begin{eqnarray} \sup\limits_{x\geq 1}x^{-1/\theta}\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}EY_{ni}\bigg| &\leq& C\sup\limits_{x\geq 1}x^{-1/\theta}\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Ea_{ni}X_{ni}I\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\bigg|\\ &&+ C\sup\limits_{x\geq 1}x^{-1/\theta}\sum\limits_{i = 1}^{n}x^{1/\theta}P\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\\ &\leq& C\sup\limits_{x\geq 1}x^{-1/\theta}\sum\limits_{i = 1}^{n}E\big|a_{ni}X_{ni}\big|I\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\\ &\leq& C\sup\limits_{x\geq 1}x^{-s/\theta}\sum\limits_{i = 1}^{n}E\big|a_{ni}X\big|^{s}I\big(\big|a_{ni}X\big|>x^{1/\theta}\big)\\ &\leq& C\sup\limits_{i\geq 1}\big|a_{ni}\big|^{s-\theta}\sum\limits_{i = 1}^{n}\big|a_{ni}\big|^{\theta}E|X|^{s} I\big(\big|a_{ni}X\big|>1\big)\\ &\leq& Cn^{\alpha-\big(s-\theta\big)r}E|X|^{s}I\big(|X|>n^{r}\big)\\ &\leq& Cn^{-1-\beta}E|X|^{s}I\big(|X|>n^{r}\big)\rightarrow 0, \; n\rightarrow\infty, \end{eqnarray} $

这样, (3.20)式仍成立.

取某个$ p $使得

(3.24) $ \begin{equation} p>\max\left\{2, \theta+\frac{1+\alpha+\beta}{r}, \frac{2\big(1+\beta\big)}{r\big(2-\theta\big)-\alpha}\right\}. \end{equation} $

从而由(3.20)式, Markov不等式, 引理2.2以及$ C_{r} $不等式, 有

(3.25) $ \begin{eqnarray} J_{2} & = &\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}Y_{ni}\bigg| >x^{1/ \theta}\bigg){\rm d}x\\ &\leq& \sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}P\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}\big(Y_{ni}-EY_{ni}\big)\bigg|>\frac {x^{1/ \theta}}{2}\bigg){\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-p/\theta}E\bigg(\max\limits_{1\leq j\leq n}\bigg|\sum\limits_{i = 1}^{j}\big(Y_{ni}-EY_{ni}\big)\bigg|^{p}\bigg){\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/\theta}E\big|Y_{ni}-EY_{ni}\big|^p{\rm d}x\\ &&+C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-p/\theta}\bigg(\sum\limits_{i = 1}^nE\big(Y_{ni}-EY_{ni}\big)^2\bigg)^{p/2}{\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}x^{-p/\theta}E\big|Y_{ni}\big|^p{\rm d}x +C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-p/\theta}\bigg(\sum\limits_{i = 1}^nEY_{ni}^2\bigg)^{p/2}{\rm d}x\\ &\triangleq& J_{2}^{(1)}+J_{2}^{(2)}. \end{eqnarray} $

根据$ p>s = \theta+\frac{1+\alpha+\beta}{r}>\theta $, 以及当$ 1+\alpha+\beta = 0 $时, 情况1中$ J_{2}<\infty $的证明过程, 容易得到$ J_{2}^{(1)}<\infty $.又根据$ Y_{ni} $的定义, 引理2.3和$ C_{r} $不等式可得

(3.26) $ \begin{eqnarray} J_{2}^{(2)} & = &C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-p/\theta}\bigg(\sum\limits_{i = 1}^nEY_{ni}^2\bigg)^{p/2}{\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-p/\theta}\bigg(\sum\limits_{i = 1}^nE\big|a_{ni}X_{ni}\big|^{2} I\big(\big|a_{ni}X_{ni}\big|\leq x^{1/\theta}\big) \bigg)^{p/2}{\rm d}x\\ &&+C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}\bigg(\sum\limits_{i = 1}^nP\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\bigg) ^{p/2}{\rm d}x\\ &\triangleq& J_{21}^{(2)}+J_{22}^{(2)}. \end{eqnarray} $

首先证明:$ J_{21}^{(2)}<\infty $.当$ s\geq2 $时, 由$ p>\frac{2(1+\beta)}{r(2-\theta)-\alpha} $以及$ E|X|^{2}<\infty $可得

(3.27) $ \begin{eqnarray} J_{21}^{(2)} &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\bigg(\int_{1}^{\infty}x^{-p/\theta}{\rm d}x\bigg)\bigg(\sum\limits_{i = 1}^n\big|a_{ni}\big|^2 E|X|^{2}\bigg)^{p/2}\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\bigg(\sum\limits_{i = 1}^n\big|a_{ni}\big|^{\theta} \sup\limits_{i\geq1}\big|a_{ni}\big|^{2-\theta}\bigg)^{p/2}\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta+\left[\alpha-r(2-\theta)\right]p/2}<\infty. \end{eqnarray} $

当$ 1\leq s<2 $时, 由$ p>2 $, $ E|X|^s<\infty $可得

(3.28) $ \begin{eqnarray} J_{21}^{(2)} &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}\bigg(\sum\limits_{i = 1}^nE\big|x^{-1/\theta}a_{ni}X_{ni}\big|^2 I\big(\big|a_{ni}X_{ni}\big|\leq x^{1/\theta}\big)\bigg)^{p/2}{\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}\bigg(\sum\limits_{i = 1}^nE\big|x^{-1/\theta}a_{ni}X_{ni}\big|^s I\big(\big|a_{ni}X_{ni}\big|\leq x^{1/\theta}\big)\bigg)^{p/2}{\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\int_{1}^{\infty}x^{-sp/2\theta}\bigg(\sum\limits_{i = 1}^nE\big|a_{ni}X_{ni}\big|^s I\big(\big|a_{ni}X_{ni}\big|\leq x^{1/\theta}\big)\bigg)^{p/2}{\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\bigg(\int_{1}^{\infty}x^{-sp/2\theta}{\rm d}x\bigg)\bigg(\sum\limits_{i = 1}^n\big|a_{ni}\big|^s E|X|^s\bigg)^{p/2}\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\bigg(\sum\limits_{i = 1}^n\big|a_{ni}\big|^\theta\sup\limits_{i\geq1}\big|a_{ni}\big|^{s-\theta}\bigg) ^{p/2}\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta+(r\theta+\alpha-rs)p/2} = C\sum\limits_{n = 1}^\infty n^{\beta-(1+\beta)p/2}<\infty. \end{eqnarray} $

其次证明:$ J_{22}^{(2)}<\infty $.注意到

(3.29) $ \begin{eqnarray} \sup\limits_{x\geq1}\bigg|\sum\limits_{i = 1}^nP\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)\bigg| &\leq& C\sum\limits_{i = 1}^nP\big(\big|a_{ni}X\big|>1\big)\\ &\leq& C\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^sI\big(\big|a_{ni}X\big|>1\big)\\ &\leq& C\sup\limits_{i\geq1}\big|a_{ni}\big|^{s-\theta}\sum\limits_{i = 1}^n\big|a_{ni}\big|^{\theta}E|X|^s I\big(\big|a_{ni}X\big|>1\big)\\ &\leq& Cn^{\alpha-(s-\theta)r}E|X|^sI\big(|X|>n^{r}\big)\\ &\leq& Cn^{-1-\beta}E|X|^sI\big(|X|>n^{r}\big)\rightarrow 0, \; n\rightarrow\infty. \end{eqnarray} $

因此, 当$ n $充分大时, $ \sum\limits_{i = 1}^nP\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big)<1 $, $ x\geq1 $.从而由(3.8)和(3.22)式可知

(3.30) $ \begin{eqnarray} J_{22}^{(2)} &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^n\int_{1}^{\infty}P\big(\big|a_{ni}X_{ni}\big|>x^{1/\theta}\big){\rm d}x\\ &\leq& C\sum\limits_{n = 1}^\infty n^{\beta}\sum\limits_{i = 1}^nE\big|a_{ni}X\big|^{\theta}I\big(\big|a_{ni}X\big|>1\big)\\ &\leq& CE|X|^{\theta+(1+\alpha+\beta)/r}<\infty. \end{eqnarray} $

情况2  $ s<1 $

取$ p>0 $, 使得$ \theta+(1+\alpha+\beta)/r = s<p<1 $, 则$ \alpha+\beta-r(p-\theta)<-1 $.类似$ 1+\alpha+\beta = 0 $中情况1的证明, 容易得到: $ J_{21}^{*}\leq CE|X|^{\theta+(1+\alpha+\beta)/r}<\infty $, $ J_{21}^{**}\leq CE|X|^{\theta+(1+\alpha+\beta)/r}<\infty $, $ J_{22}\leq CE|X|^{\theta+(1+\alpha+\beta)/r}<\infty $以及$ J_{23}\leq CE|X|^{\theta+(1+\alpha+\beta)/r}<\infty $.证毕.