Borel–Cantelli Lemma for Capacities

Abstract

In this paper, we investigate the second Borel–Cantelli lemma for capacity without the assumption of independence for events. We obtain a sufficient condition under which the second Borel–Cantelli lemma for capacity holds. Our results are natural extensions of the classical Borel–Cantelli lemma. However, the proof is different from the existing literature.

1. Introduction

In probability theory, the Borel–Cantelli lemma, which was first obtained by Borel (1909, 1912) [1,2] and Cantelli (1917) [3], is an important theorem about sequences of events, such a lemma consists of two parts, which are called the first and second Borel–Cantelli lemma. The lemma states that, under certain conditions, an event will occur with either probability zero or probability one. The second Borel–Cantelli lemma is a partial converse of the first Borel–Cantelli lemma since the second Borel–Cantelli lemma needs the additional assumption of independence.

Since then, there has been a large amount of literature to extend the Borel–Cantelli lemma (see, for example, Chandra (2012) [4] in detail). For the first Borel–Cantelli lemma, we refer the reader to Barndorff-Nielsen (1961) [5], Martikainen and Petrov (1990) [6], and Balakrishnan and Stepanov (2010) [7]. For the second Borel–Cantelli lemma, we refer the reader to Erdös and Rényi (1959) [8], Kochen and Stone (1964) [9], Petrov (2002, 2004) [10,11], Yan (2006) [12], (which provides a new and simple proof about Kochen and Stone (1964) [9]), and Xie (2009) [13] and Zong et al. (2016) [14], in which many attempts can be founded to weaken the independence condition.

Recently, motivated by mathematical finance and robust statistics, non-additive measures or non-linear expectations have already caught many scholars’ attention; see Huber (1973) [15], Denneberg (1994) [16], Wang and Klir (2009) [17], Peng (2006, 2009, 2019) [18,19,20], Torra et al. (2014) [21], and Zong et al. (2016) [14]. A natural question is whether or not such a Borel–Cantelli lemma could be extended to the case where the probability measure is non-additive.

According to the property of the first Borel–Cantelli lemma, we can easily extend the lemma to the case where the probability is no longer additive. In fact, the first Borel–Cantelli lemma holds for all set functions that are of countable subadditivity and monotonicity; see Billingsley (1995) [22]. However, for the second Borel–Cantelli lemma, it is not easy to do so because this lemma depends on the assumption of independence. Therefore, many concepts of independence under nonadditive probability/expectation have been introduced, for example, Peng’s independence in Peng (2006, 2009, 2019) [18,19,20], Marinacci pre-independence in Maccheroni and Marinacci (2005) [23], and Puhalskii independence (2001) [24], as well as the Fubini-Like Theorem for Choquet Integrals, in Zong et al. (2016) [14]. A natural question is whether we could investigate the second Borel–Cantelli lemma for capacities without the assumption of independence. In this paper, we obtain a sufficient condition under which the second Borel–Cantelli lemma for capacity holds. It turns out that our results are natural extensions of the classical Borel–Cantelli lemma. However, the proofs are different from the existing literature.

This paper is organized as follows: In Section 2, we show some basic definitions and propositions with respect to capacity and present some preparatory lemmas. In Section 3, we provide a sufficient condition under which the second Borel–Cantelli lemma for capacities holds. In Section 4, we consider the case where random variables are independent under non-additive expectations or non-additive probabilities.

2. Preliminaries

Assume that $(\mathrm{\Omega },\mathcal{F})$ is a measurable space; we define capacity, V, as follows.

Definition 1. 

A set-function, V, on $\mathcal{F}$ is called a capacity if it satisfies

(i) 

$V(\varnothing )=0,V\left(\mathrm{\Omega }\right)=1.$

(ii) 

$V\left(A\right)\le V\left(B\right),$  $A\subset B,$  $A,B\in \mathcal{F}.$

(iii) 

$V(A\bigcup B)\le V\left(A\right)+V\left(B\right),$$A,B\in \mathcal{F}.$

Given a capacity, V, let the $\mathcal{F}$-measurable function $X:\mathrm{\Omega }\to R$ be a random variable defined on $(\mathrm{\Omega },\mathcal{F}).$ We focus on Choquet expectation, $\mathbb{E}$, denoted as

Definition 2. 

Given a capacity, V, a Choquet (integral) expectation is denoted as

$$\mathbb{E}\left[X\right]:={\int }_0^{\infty }V(X\ge t)dt+{\int }_{-\infty }^0\left[V(X>t)-1\right]dt.$$

We assume that $\mathcal{H}$ is the set of all random variables, X, with $\mathbb{E}\left[\right|X\left|\right]<\infty .$

Definition 3. 

Two random variables, $\xi ,\eta \in \mathcal{H}$, are comonotone if, almost surely,

$$(\xi \left(\omega \right)-\xi \left({\omega }^{\prime }\right))(\eta \left(\omega \right)-\eta \left({\omega }^{\prime }\right))\ge 0.$$

For more knowledge about comonotonicity, see, for instance, Dhaene et al. (2002) [25].

The basic properties of Choquet expectations are given in the following proposition (see, e.g., Denneberg (1992) [16]).

Lemma 1. 

(a) 

Monotonicity: $X,Y\in \mathcal{H},$ if $X\ge Y$, then $\mathbb{E}\left[X\right]\ge \mathbb{E}\left[Y\right]$.

(b) 

Constant preserving: $\mathbb{E}\left[c\right]=c,\forall c\in \mathbb{R}$.

(c) 

Translation invariance: $\mathbb{E}[c+X]=c+\mathbb{E}\left[X\right],\forall c\in \mathbb{R}$.

(d) 

Positive homogeneity: $\mathbb{E}\left[\lambda X\right]=\lambda \mathbb{E}\left[X\right],\forall \lambda \ge 0$.

(e) 

Lower–upper Choquet expectations: $-\mathbb{E}[-X]\le \mathbb{E}\left[X\right].$

(f) 

Comonotonic additivity: if $X,Y$ are comonotonic random variables, then

$$\mathbb{E}[X+Y]=\mathbb{E}\left[X\right]+\mathbb{E}\left[Y\right].$$

Remark 1. 

Usually, a Choquet expectation does not satisfy the following sub-linearity:

$$\mathbb{E}[X+Y]\le \mathbb{E}\left[X\right]+\mathbb{E}\left[Y\right].$$

However, it has been proven that a Choquet expectation satisfies sub-linearity if and only if the corresponding capacity, V, is $2$—alternating in the sense of

$$V(A\cup B)\le V\left(A\right)+V\left(B\right)-V(A\cap B),A,B\in \mathcal{F}.$$

Proposition 1. 

Let φ on $\mathbb{R}$ be a positively convex function. Then, the Jensen inequality under Choquet expectation holds:

$$\phi \left(\mathbb{E}\right[X\left]\right)\le \mathbb{E}\left[\phi \right(X\left)\right].$$

Proof. 

First, it is easy to check that a Choquet expectation has the following property:

$$\lambda \mathbb{E}\left[X\right]\le \mathbb{E}\left[\lambda X\right],\lambda \in \mathbb{R}.$$

In fact, obviously, the above inequality becomes equality if $\lambda \ge 0$, by the definition of Choquet expectation. We now prove the case where $\lambda \le 0.$ In fact,

$$\lambda \mathbb{E}\left[X\right]=-\left|\lambda \right|\mathbb{E}\left[X\right]=-\mathbb{E}\left[\right|\lambda \left|X\right]=-\mathbb{E}[-\lambda X]\le \mathbb{E}\left[\lambda X\right].$$

The last inequality is due to Lemma 1(e).

Using the above inequality, we can easily prove this lemma. Indeed, because the function $\phi :\mathbb{R}\to \mathbb{R}$ is convex, there exists a countable set, D, in ${\mathbb{R}}^2$, such that $\phi \left(x\right)=\underset{(a,b)\in D}{sup}(ax+b).$ Via the translation invariance of the Choquet expectation in Lemma 1(c), we have

$$\begin{array}{ccc}\hfill \phi \left(\mathbb{E}\right[X\left]\right)& =& \underset{(a,b)\in D}{sup}(a\mathbb{E}\left[X\right]+b)\hfill \\ & \le & \underset{(a,b)\in D}{sup}\left(\mathbb{E}[aX+b]\right)\hfill \\ & \le & \mathbb{E}\left[\underset{(a,b)\in D}{sup}(aX+b)\right]\hfill \\ & =& \mathbb{E}\left[\phi \right(X\left)\right].\hfill \end{array}$$

The proof is complete. □

Because the probability measure in probability theory is assumed to be continuous in the sense that $P\left(A_n\right)\to P\left(A\right)$ whenever $A_n\to A,n\to \infty$, Fatou’s lemma is naturally true. However, for capacities, Fatou’s lemma is usually not true because the capacity in the nonlinear case is no longer continuous. Thus, we need the following concept.

Definition 4. 

A capacity, V, is called a Fatou-like capacity if

$$V\left(\underset{n\to \infty }{lim\; sup}{A}_n\right)\ge \underset{n\to \infty }{lim\; sup}V\left(A_n\right),A_n\in \mathcal{F}.$$

(1)

It is easy to show that the following capacities are Fatou-like capacities:

Example 1. 

Let $B_n,B\in \mathcal{F}$ and $\underset{n\to \infty }{lim}V\left(B_n\right)=V\left(B\right)$ whenever $B_n\downarrow B,$ and then V is a Fatou-like capacity.

Example 2. 

Let $\mathcal{P}$ be a weakly compact set of probability measures defined on $(\mathrm{\Omega },\mathcal{F})$; then, the upper probability, V, defined by

$$V\left(A\right)=\underset{Q\in \mathcal{P}}{sup}Q\left(A\right),$$

is a Fatou-like capacity.

The following lemma is an important lemma that we use in this paper. The main idea is from Yan (2006) [12].

Lemma 2. 

Let X be a random variable, such that $\mathbb{E}\left[e^X\right]>e^{\alpha }$ for a constant, $\alpha ,$ and then

$$V(X>\alpha )\ge {\displaystyle \frac{{\left(\mathbb{E}\left[e^X\right]-e^{\alpha }\right)}^2}{\mathbb{E}\left[{\left(e^X-e^{\alpha }\right)}^2\right]}}.$$

Proof. 

It is easy to check that, for any $x,\alpha \in R,$

$$\left(e^x-e^{\alpha }\right)I_{\{x>\alpha \}}\ge e^x-e^{\alpha },$$

here and in the sequel, $I_A$ represents the indicator function of set $A.$

Therefore,

$$\mathbb{E}\left[(e^X-e^{\alpha })I_{\{X>\alpha \}}\right]\ge \mathbb{E}\left[e^X\right]-e^{\alpha }\ge 0.$$

(2)

For convenience, we denote

$$\xi :=(e^X-e^{\alpha })I_{\{X>\alpha \}}\mathrm{and}\eta :=I_{\{X>\alpha \}}.$$

It then follows the elementary inequality $\left|ab\right|\le {\displaystyle \frac{1}{2}}{a}^2+{\displaystyle \frac{1}{2}}{b}^2$ that

$${\displaystyle \frac{\left|\xi \eta \right|}{{\left(\mathbb{E}\left[\right|\xi {|}^2]\right)}^{\frac{1}{2}}{\left(\mathbb{E}\left[\right|\eta {|}^2]\right)}^{\frac{1}{2}}}}\le {\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\xi \right|}^2}{\mathbb{E}\left[\right|\xi {|}^2]}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\eta \right|}^2}{\mathbb{E}\left[\right|\eta {|}^2]}},$$

(3)

Taking the Choquet integration $\mathbb{E}[·]$ on both sides in inequality (3), according to the monotonicity of Choquet integration in Lemma 1(a), we have

$$\mathbb{E}\left[{\displaystyle \frac{\left|\xi \eta \right|}{{\left(\mathbb{E}\left[\right|\xi {|}^2]\right)}^{\frac{1}{2}}{\left(\mathbb{E}\left[\right|\eta {|}^2]\right)}^{\frac{1}{2}}}}\right]\le \mathbb{E}\left[{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\xi \right|}^2}{\mathbb{E}\left[\right|\xi {|}^2]}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\eta \right|}^2}{\mathbb{E}\left[\right|\eta {|}^2]}}\right].$$

(4)

Furthermore, it is easy to check that both ${\left|\xi \right|}^2={(e^X-e^{\alpha })}^2{I}_{\{X>\alpha \}}$ and ${\left|\eta \right|}^2=I_{\{X>\alpha \}}$ are co-monotonic; hence, via the comonotonic additivity of the Choquet expectation, we get

$$\mathbb{E}\left[{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\xi \right|}^2}{\mathbb{E}\left[\right|\xi {|}^2]}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\eta \right|}^2}{\mathbb{E}\left[\right|\eta {|}^2]}}\right]=\mathbb{E}\left[{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\xi \right|}^2}{\mathbb{E}\left[\right|\xi {|}^2]}}\right]+\mathbb{E}\left[{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\eta \right|}^2}{\mathbb{E}\left[\right|\eta {|}^2]}}\right]=1$$

This, with (4), implies that

$${\left(\mathbb{E}\right[\left|\xi \eta \right|\left]\right)}^2\le {\mathbb{E}\left[\right|\xi |}^2{\left]\mathbb{E}\right[\left|\eta \right|}^2].$$

That is

$$\begin{array}{ccc}\hfill {\left(\mathbb{E}\left[\left|(e^X-e^{\alpha })I_{\{X>\alpha \}}\right|\right]\right)}^2& \le & \mathbb{E}\left[\right|(e^X-e^{\alpha })I_{\{X>\alpha \}}{|}^2\left]\mathbb{E}\right[|I_{\{X>\alpha \}}{|}^2]\hfill \\ & \le & \mathbb{E}\left[\right|(e^X-e^{\alpha }){|}^2]V(X>\alpha ),\hfill \end{array}$$

which, with (2), implies the desired result. □

Lemma 3. 

Suppose that ${\left\{A_i\right\}}_{i=1}^{\infty }$ is a sequence of events; set $X_i:=I_{A_i}.$ Assume that

$$a_n:=\sum _{i=1}^{n}V\left(A_i\right)\to \infty ,n\to \infty ,$$

and for any sequence of real numbers, $\left\{k_n\right\}$ with $k_n>0,$

$$\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{\mathbb{E}\left[e^{k_n{\sum }_{i=1}^n{X}_i}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{k_n{X}_i}\right]}}<\infty .$$

Then,

(I) 

For any constant, $\alpha >0$, and any sequence, $\left\{b_n\right\}$ with $b_n\ge \sqrt{a_n},$

$$\underset{n\ge 1}{sup}\mathbb{E}\left[e^{{\displaystyle \frac{\alpha }{b_n}}{\sum }_{i=1}^n\left(X_i-\mathbb{E}\left[X_i\right]\right)}\right]<\infty .$$

(5)

(II) 

For any $ϵ>0,$

$$\underset{n\to \infty }{lim}V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>1+ϵ\right)=0.$$

(6)

(III) 

For any constant $\alpha >0,$

$$\underset{n\to \infty }{lim}\mathbb{E}\left[e^{{\displaystyle \frac{\alpha }{a_n}}{\sum }_{i=1}^n{X}_i}{I}_{\{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i\ge 1+\epsilon \}}\right]=0.$$

(7)

Proof. 

The proof of (I):

A trite calculation of a double integral, the following elementary equality can be obtained easily:

$$e^x=1+x+x^2{\int }_0^{1}rdr{\int }_0^1{e}^{rxy}dy,x\in (-\infty ,+\infty ).$$

Immediately, we get

$$e^x\le 1+x+{\displaystyle \frac{x^2}{2}}{e}^{\left|x\right|},x\in (-\infty ,+\infty ).$$

(8)

Choosing $x={\displaystyle \frac{\alpha }{b_n}}(X_i-\mathbb{E}\left[X_i\right])$ in (8).

Since $|X_i-\mathbb{E}\left[X_i\right]|=|I_{A_i}-V\left(A_i\right)|\le 2$ and $b_n\ge \sqrt{a_n}\to \infty$ as $n\to \infty ,$ thus, for a sufficiently large $n,$ we have $b_n\ge 1$

$${\displaystyle \frac{\alpha }{b_n}}|X_i-\mathbb{E}\left[X_i\right]|\le {\displaystyle \frac{2\alpha }{b_n}}\le 2\alpha .$$

Note the fact that $X_i^2=X_i.$ Thus,

$$\begin{array}{ccc}\hfill e^{{\displaystyle \frac{\alpha }{b_n}}(X_i-\mathbb{E}\left[X_i\right])}& \le & 1+{\displaystyle \frac{\alpha }{b_n}}(X_i-\mathbb{E}\left[X_i\right])+{\displaystyle \frac{{\alpha }^2{|X_i-\mathbb{E}\left[X_i\right]|}^2}{2b_n^2}}{e}^{{\displaystyle \frac{\alpha }{b_n}}|X_i-\mathbb{E}\left[X_i\right]|}\hfill \\ & \le & 1+{\displaystyle \frac{\alpha }{b_n}}(X_i-\mathbb{E}\left[X_i\right])+{\displaystyle \frac{{\alpha }^2{|X_i-\mathbb{E}\left[X_i\right]|}^2}{2a_n}}{e}^{2\alpha }\hfill \\ & \le & 1+{\displaystyle \frac{\alpha }{b_n}}(X_i-\mathbb{E}\left[X_i\right])+{\displaystyle \frac{{\alpha }^2{e}^{2\alpha }}{2a_n}}\left((1+2\mathbb{E}\left[X_i\right])+\mathbb{E}{\left[X_i\right]}^2\right)X_i\hfill \\ & =& 1+\left({\displaystyle \frac{\alpha }{b_n}}+{\displaystyle \frac{{\alpha }^2{e}^{2\alpha }}{2a_n}}(1+2\mathbb{E}\left[X_i\right])\right)X_i-{\displaystyle \frac{\alpha }{b_n}}\mathbb{E}\left[X_i\right]+{\displaystyle \frac{{\alpha }^2{e}^{2\alpha }}{2a_n}}\mathbb{E}{\left[X_i\right]}^2\hfill \end{array}$$

Set expectation $\mathbb{E}[·]$ on both sides of the inequality above; given the translation invariance in Lemma 1 and $\mathbb{E}\left[X_i\right]=V\left(A_i\right)\le 1,$ immediately,

$$\begin{array}{ccc}\hfill \mathbb{E}\left[e^{{\displaystyle \frac{\alpha }{b_n}}(X_i-\mathbb{E}\left[X_i\right])}\right]& \le & 1+\left({\displaystyle \frac{{\alpha }^2{e}^{2\alpha }}{2a_n}}(1+2\mathbb{E}\left[X_i\right])\right)\mathbb{E}\left[X_i\right]+{\displaystyle \frac{{\alpha }^2{e}^{2\alpha }}{2a_n}}{\left(\mathbb{E}\left[X_i\right]\right)}^2\hfill \\ & \le & 1+{\displaystyle \frac{4{\alpha }^2{e}^{2\alpha }}{2a_n}}V\left(A_i\right)\hfill \\ & \le & e^{\left({\displaystyle \frac{2{\alpha }^2{e}^{2\alpha }}{a_n}}V\left(A_i\right)\right)},\hfill \end{array}$$

The last inequality follows from the fact that, for $x\ge 0,$$1+x\le e^x.$

Therefore, for a sufficiently large $n,$ we have

$$\begin{array}{ccc}\hfill \mathbb{E}\left[e^{{\displaystyle \frac{\alpha }{b_n}}{\sum }_{i=1}^n(X_i-\mathbb{E}\left[X_i\right])}\right]& =& {\displaystyle \frac{\mathbb{E}\left[e^{\frac{\alpha }{b_n}{\sum }_{i=1}^n{X}_i}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{\frac{\alpha }{b_n}{X}_i}\right]}}\prod _{i=1}^n\mathbb{E}\left[e^{{\displaystyle \frac{\alpha }{b_n}}(X_i-\mathbb{E}\left[X_i\right])}\right]\hfill \\ & \le & {\displaystyle \frac{\mathbb{E}\left[e^{\frac{\alpha }{b_n}{\sum }_{i=1}^n{X}_i}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{\frac{\alpha }{b_n}{X}_i}\right]}}{e}^{\left({\displaystyle \frac{4{\alpha }^2{e}^{2\alpha }}{a_n}}{\sum }_{i=1}^{n}V\left(A_i\right)\right)}\hfill \\ & =& e^{4{\alpha }^2{e}^{2\alpha }}\underset{n\ge 1}{sup}{\displaystyle \frac{\mathbb{E}\left[e^{\frac{\alpha }{b_n}{\sum }_{i=1}^n{X}_i}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{\frac{\alpha }{b_n}{X}_i}\right]}}<\infty .\hfill \end{array}$$

The proof of (I) is complete.

The proof of (II): For any $ϵ>0,$ via Markov’s inequality, we get

$$\begin{array}{ccc}& & V\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{X}_i\ge 1+ϵ\right)\hfill \\ & =& V\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}(X_i-\mathbb{E}\left[X_i\right])\ge ϵ\right)\hfill \\ & =& V\left({\displaystyle \frac{1}{\sqrt{a_n}}}{\displaystyle \sum _{i=1}^n}(X_i-\mathbb{E}\left[X_i\right])\ge ϵ\sqrt{a_n}\right)\hfill \\ & \le & e^{-ϵ\sqrt{a_n}}\mathbb{E}\left[exp\left({\displaystyle \frac{1}{\sqrt{a_n}}}{\displaystyle \sum _{i=1}^n}(X_i-\mathbb{E}\left[X_i\right])\right)\right]\to 0,n\to \infty \hfill \end{array}$$

due to (I) and $a_n\to \infty ,n\to \infty .$

The proof of (III): Let $\xi :=e^{{\displaystyle \frac{\alpha }{a_n}}{\sum }_{i=1}^n{X}_i}$ and $\eta :=I_{\{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i\ge 1+\epsilon \}}$. It is easy to obtain confirmation that $\xi$ and $\eta$ are comonotonic. Similarly to (3), we have

$$\mathbb{E}\left[{\displaystyle \frac{\left|\xi \eta \right|}{{\left(\mathbb{E}\left[\right|\xi {|}^2]\right)}^{\frac{1}{2}}{\left(\mathbb{E}\left[\right|\eta {|}^2]\right)}^{\frac{1}{2}}}}\right]\le \mathbb{E}\left[{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\xi \right|}^2}{\mathbb{E}\left[\right|\xi {|}^2]}}+{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left|\eta \right|}^2}{\mathbb{E}\left[\right|\eta {|}^2]}}\right]=1,$$

due to the comonotonic additivity of Choquet integration. Thereby, we obtain

$$\begin{array}{ccc}& & {\left(\mathbb{E}\left[e^{{\displaystyle \frac{\alpha }{a_n}}{\sum }_{i=1}^n{X}_i}{I}_{\{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i\ge 1+\epsilon \}}\right]\right)}^2\hfill \\ & \le & \mathbb{E}\left[e^{{\displaystyle \frac{2\alpha }{a_n}}{\sum }_{i=1}^n{X}_i}\right]V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i\ge 1+\epsilon \right)\hfill \\ & =& e^{2\alpha }\mathbb{E}\left[e^{{\displaystyle \frac{2\alpha }{a_n}}{\sum }_{i=1}^n(X_i-\mathbb{E}\left[X_i\right])}\right]V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i\ge 1+\epsilon \right)\to 0\hfill \end{array}$$

due to (I) and (II). □

3. The Second Borel–Cantelli Lemma for Capacities

We now begin to prove the second Borel–Cantelli lemma for capacities:

Theorem 1. 

Let V be a Fatou-like capacity, and let ${\left\{A_n\right\}}_{n=1}^{\infty }$ be a sequence of events, such that

$$\sum _{i=1}^{\infty }V\left(A_i\right)=\infty .$$

If, for any sequence $k_n>0,$

$$1\le \underset{n\to \infty }{lim\; inf}{\displaystyle \frac{\mathbb{E}\left[e^{k_n{\sum }_{i=1}^n{I}_{A_i}}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{k_n{I}_{A_i}}\right]}}\le \underset{n\to \infty }{lim\; sup}{\displaystyle \frac{\mathbb{E}\left[e^{k_n{\sum }_{i=1}^n{I}_{A_i}}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{k_n{I}_{A_i}}\right]}}<\infty ,$$

(9)

then

$$V(A_{n}i.o.)=1.$$

Proof. 

Set $X_i=I_{A_i}$ and $a_n:={\displaystyle \sum _{i=1}^n}\mathbb{E}\left[X_i\right]$. Immediately, $\mathbb{E}\left[X_i\right]=V\left(A_i\right)$, and

$$a_n=\sum _{i=1}^n\mathbb{E}\left[X_i\right]=\sum _{i=1}^{n}V\left(A_i\right)\to \infty .$$

Now, we consider the following two events:

$$\{\omega :\underset{n\to \infty }{lim\; sup}{A}_n\}\mathrm{and}\{\omega :\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>\epsilon \},$$

and here, $\epsilon \in (0,1).$

If $\omega \notin \left\{\underset{n\to \infty }{lim\; sup}{A}_n\right\}$ holds, then $\underset{n\to \infty }{lim}{\sum }_{i=1}^n{X}_i=\underset{n\to \infty }{lim}{\sum }_{i=1}^n{I}_{A_i}$ is a finite number. Hence,

$${\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i\to 0\mathrm{as}n\to \infty$$

because of $a_n\to \infty .$ This implies $\omega \notin \{\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i>\epsilon \}.$ Therefore, we have the following inclusion relation:

$$\{\omega :\underset{n\to \infty }{lim\; sup}{A}_n\}\supseteq \{\omega :\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>\epsilon \}.$$

The monotonicity of the capacity V and the definition of Fatou-like capacity (1) imply that

$$V\left(\underset{n\to \infty }{lim\; sup}{A}_n\right)\ge V\left(\underset{n\to \infty }{lim\; sup}{\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>\epsilon \right)\ge \underset{n\to \infty }{lim\; sup}V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>\epsilon \right).$$

(10)

In order to apply Lemma 2, we need to check the conditions of Lemma 2.

Given the assumption of Theorem that

$$\underset{n\to \infty }{lim\; inf}{\displaystyle \frac{\mathbb{E}\left[e^{\frac{1}{a_n}{\sum }_{i=1}^n{X}_i}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{\frac{1}{a_n}{X}_i}\right]}}\ge 1$$

and Jensen’s inequality, we have, for $\epsilon \in (0,1),$

$$\begin{array}{ccc}\hfill \mathbb{E}\left[e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}\right]-e^{\epsilon }& =& {\displaystyle \frac{\mathbb{E}\left[e^{\frac{1}{a_n}{\sum }_{i=1}^n{X}_i}\right]}{{\prod }_{i=1}^n\mathbb{E}\left[e^{\frac{1}{a_n}{X}_i}\right]}}\prod _{i=1}^n\mathbb{E}\left[e^{{\displaystyle \frac{1}{a_n}}{X}_i}\right]-e^{\epsilon }\hfill \\ & \ge & \prod _{i=1}^n{e}^{{\displaystyle \frac{1}{a_n}}\mathbb{E}\left[X_i\right]}-e^{\epsilon }=e-e^{\epsilon }>0.\hfill \end{array}$$

Thanks to Lemma 2, we thus have

$$V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>\epsilon \right)\ge {\displaystyle \frac{{\left(\mathbb{E}\left[e^{\frac{1}{a_n}{\sum }_{i=1}^n{X}_i}\right]-e^{\epsilon }\right)}^2}{\mathbb{E}\left[{\left(e^{\frac{1}{a_n}{\sum }_{i=1}^n{X}_i}-e^{\epsilon }\right)}^2\right]}}.$$

(11)

For the numerator in fraction (11), we see that

$${\left(\mathbb{E}\left[e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}\right]-e^{\epsilon }\right)}^2\ge {(e-e^{\epsilon })}^2.$$

(12)

On the other hand, for the denominator in fraction (11), note that $X_i\ge 0,$ and we have

$$\begin{array}{ccc}& & \mathbb{E}\left[{\left(e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}-e^{\epsilon }\right)}^2\right]\hfill \\ & =& \mathbb{E}\left[{\left(e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}-e^{\epsilon }\right)}^2{I}_{\left\{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i\le 1+\epsilon \right\}}+{\left(e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}-e^{\epsilon }\right)}^2{I}_{\left\{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i>1+\epsilon \right\}}\right]\hfill \\ & \le & {\left(e^{1+\epsilon }-e^{\epsilon }\right)}^2+\mathbb{E}\left[{\left(e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}-e^{\epsilon }\right)}^2{I}_{\left\{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i>1+\epsilon \right\}}\right].\hfill \end{array}$$

For the second term, for simplicity, we write $B_n:=\left\{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i>1+\epsilon \right\}$; via the comonotonic additivity of Choquet expectation, we have

$$\begin{array}{ccc}& & \mathbb{E}\left[{\left(e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}-e^{\epsilon }\right)}^2{I}_{B_n}\right]\hfill \\ & \le & \mathbb{E}\left[e^{{\displaystyle \frac{2}{a_n}}{\sum }_{i=1}^n{X}_i}{I}_{B_n}+2e^{\epsilon }{e}^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}{I}_{B_n}+e^{2\epsilon }{I}_{B_n}\right]\hfill \\ & =& \mathbb{E}\left[e^{{\displaystyle \frac{2}{a_n}}{\sum }_{i=1}^n{X}_i}{I}_{B_n}\right]+2e^{\epsilon }\mathbb{E}\left[e^{{\displaystyle \frac{1}{a_n}}{\sum }_{i=1}^n{X}_i}{I}_{B_n}\right]+e^{2\epsilon }V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>1+\epsilon \right).\hfill \end{array}$$

According to (III) in Lemma 3, the first term and the second term on the right-hand side go to zero as $n\to \infty .$ The last term on the right-hand side also goes to zero as $n\to \infty$, according to (II) in Lemma 3.

This, with (11) and (12), implies that

$$\underset{n\to \infty }{lim\; sup}V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>0\right)\ge \underset{n\to \infty }{lim\; sup}V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>\epsilon \right)\ge {\displaystyle \frac{{(e-e^{\epsilon })}^2}{{\left(e^{1+\epsilon }-e^{\epsilon }\right)}^2}}.$$

(13)

Letting $\epsilon \to 0$, we arrive at

$$\underset{n\to \infty }{lim\; sup}V\left({\displaystyle \frac{1}{a_n}}\sum _{i=1}^n{X}_i>0\right)=1.$$

(14)

According to (10) and (14), we have

$$V\left(\underset{n\to \infty }{lim\; sup}{A}_n\right)=1.$$

Therefore, we complete the proof of this theorem. □

By this theorem, immediately, we have

Corollary 1. 

Assume that a sequence of events, $\left\{A_i\right\}$, is exponentially independent under a Choquet expectation, $\mathbb{E}$, in the sense that

$$\mathbb{E}\left[e^{k_n{\sum }_{i=1}^n{I}_{A_i}}\right]=\prod _{i=1}^n\mathbb{E}\left[e^{k_n{I}_{A_i}}\right],\forall n\ge 1,\forall k_n\in R_{+}.$$

Then, Condition 9 in Theorem 1 holds.

4. Independence Cases

The notion of independence for random variables under non-additive expectation or non-additive probability is important. Motivated by mathematical finance and robust statistics, various different notions of independence have been investigated, for example, Peng’s independence, Marinacci pre-independence, and Puhalskii independence. It can be proven that all notions mentioned above satisfy Condition 9 in Theorem 1.

We now check that Condition 9 in Theorem 1 holds if event $\left\{A_i\right\}$ is Puhalskii-independent. The remaining cases can be verified in a similar manner. As defined below, Puhalskii independence implies the following:

Definition 5. 

A sequence, $\left\{A_i\right\}$, of events is said to be Puhalskii -independent if

$$V\left(\bigcap _{i=1}^n{A}_i\right)=\prod _{i=1}^{n}V\left(A_i\right).$$

This leads to the following lemma.

Lemma 4. 

Let $\left\{A_i\right\}$ be a sequence of Puhalskii-independent events. Then, for any $a_n\in R^{+},$

$$\mathbb{E}\left[e^{a_n{\sum }_{i=1}^n{I}_{A_i}}\right]=\prod _{i=1}^n\mathbb{E}\left[e^{a_n{I}_{A_i}}\right].$$

Proof. 

Let ${\phi }_i\left(x\right)=e^{a_{i}x},i=1,\cdots ,n$

$$\begin{array}{ccc}\hfill \mathbb{E}\left[\prod _{i=1}^n{\phi }_i\left(X_i\right)\right]& =& \mathbb{E}\left[{\int }_0^{\infty }\cdots {\int }_0^{\infty }\prod _{i=1}^n{I}_{\{{\phi }_i\left(X_i\right)>x_i\}}\prod _{i=1}^{n}d{x}_i\right]\hfill \\ & =& {\int }_0^{\infty }\cdots {\int }_0^{\infty }\mathbb{E}\left[\prod _{i=1}^n{I}_{\{{\phi }_i\left(X_i\right)>x_i\}}\right]\prod _{i=1}^{n}d{x}_i\hfill \\ & =& {\int }_0^{\infty }\cdots {\int }_0^{\infty }V\left(\bigcap _{i=1}^n\{{\phi }_i\left(X_i\right)>x_i\}\right)\prod _{i=1}^{n}d{x}_i\hfill \\ & =& {\int }_0^{\infty }\cdots {\int }_0^{\infty }\prod _{i=1}^{n}V\left(\{{\phi }_i\left(X_i\right)>x_i\}\right)d{x}_i\hfill \\ & =& \prod _{i=1}^n\mathbb{E}\left[{\phi }_i\left(X_i\right)\right],\hfill \end{array}$$

Here, we have used Fubini’s theorem for Choquet expectation (see Ghirardato (1997) [26] or Chateauneuf and Lefort (2008) [27]). □