Subgame perfect equilibrium in a bargaining model with deterministic procedures

Abstract

Two players, A and B, bargain to divide a perfectly divisible pie. In a bargaining model with constant discount factors, $\delta _A$ and $\delta _B$, we extend Rubinstein (Econometrica 50:97–110, 1982)’s alternating offers procedure to more general deterministic procedures, so that any player in any period can be the proposer. We show that each bargaining game with a deterministic procedure has a unique subgame perfect equilibrium (SPE) payoff outcome, which is efficient. Conversely, each efficient division of the pie can be supported as an SPE outcome by some procedure if $\delta _A+\delta _B\ge 1$, while almost no division can ever be supported in SPE if $\delta _A+\delta _B < 1$.

Appendix

Proof

(Proof of Theorem 1)

By symmetry, we only consider $\omega \in \varOmega _A$. The theorem can be easily proved if $\omega =A$, so we only consider procedures $\omega$, such that $r(w)\ge 1$.

First, define a strategy pair $S^*=(S^*_A,S^*_B)$ as follows. When the game comes to period s, for any history, the proposer $\omega _s=i$ proposes $\theta (\omega (s))$ as the share for herself and $1-\theta (\omega (s))$ as the share for her opponent $j\ne i$ according to $S_i^*$; the other player j accepts any offer in which the proposer’s share is no more than $\theta (\omega (s))$ and rejects all other offers according to $S_j^*$. It is obvious that if players follow $S^*$, then the game will end by period $t=1$ with agreement $\big (\theta (\omega ),1-\theta (\omega )\big )$. We shall prove that $S^*$ is an SPE of $G(\omega ,\delta _{A},\delta _{B})$, and $\big (\theta (\omega ),1-\theta (\omega )\big )$ is the only payoff pair that could ever be supported in SPE.

To show that $S^*$ is an SPE, it is sufficient to verify that $S^*$ satisfies the one-shot deviation principle, which asserts that a strategy profile S is an SPE of an extensive game with constant discount factors, if and only if no player can ever become strictly better off by deviating from S for just one period and then reverting to S.⁷ Suppose the game has come to period $t=s$. If the proposer $\omega _s$, without loss of generality assumed to be A, deviates from $S_A^*$ and suggests an offer in which her own share is larger than $\theta (\omega (s))$, then player B will reject this offer according to $S^*_B$, and let the game enter period $s+1$. If $\omega _{s+1}=A$, then A will get $v(s+1)=\theta (\omega (s+1))$ in period $s+1$ according to $S^*$; if $\omega _{s+1}=B$, then A will get $v(s+1)=1-\theta (\omega (s+1))$ in period $s+1$. In either case, if $\delta _B>0$, we have

$$\begin{aligned} {\begin{matrix} \theta (\omega (s))&{}=1+\sum _{k=1}^{r(\omega (s))}(-1)^k p(\omega (s),k)\\ &{}>1+\frac{1}{\delta _B}\sum _{k=1}^{r(\omega (s))}(-1)^k p(\omega (s),k)\\ &{}=v(s+1)>\delta _Av(s+1), \end{matrix}} \end{aligned}$$

where $\sum _{k=1}^{r(\omega (s))}(-1)^k p(\omega _s,k)<0$ due to Lemma 2; if $\delta _B=0$, then $\theta (\omega (s))=1$, thus we also have $\theta (\omega (s))>\delta _Av(s+1)$ due to Lemma 4. This implies that A has no incentive to increase her own share in the offer in period s, since, otherwise, her payoff will decline from $u_A(\theta (\omega (s)),s)$ to $u_A(v(s+1),s+1)$. On the other hand, it is obvious that A will also not decrease her own share in the offer in period s. Similarly, we can show that B will neither increase nor decrease her reservation share $1-\theta (\omega (s))$ for accepting an offer proposed by A in period s. Thus, we have proved that $S^*$ satisfies the one-shot deviation principle, and hence, it is an SPE.

Now, we turn to the uniqueness part of the theorem, using a technique introduced by Shaked and Sutton (1984). Given $i=0,1,\ldots ,r(\omega )$, let $\Lambda _i$ be the subset of [0, 1], so that for each $x\in \Lambda _i$, there exists an SPE $f^x$ of the game $G(\omega ,\delta _{A},\delta _{B})$ satisfying the following condition: when the game has come to period $t_i=\sum _{j=0}^i n_j+1$ (where $n_0=0$), according to $f^x$, the proposer would suggest an agreement in which his own share is x and his opponent’s share is $1-x$. Let $M_i$ and $m_i$ denote the supremum and infimum of $\Lambda _i$, respectively.

In period $n_1+1$, player B can get no more than $M_1$ in any SPE. Hence, A will offer B no more than $\delta _BM_1$ in period $n_1$, no more than $\delta _B^2 M_1$ in period $n_1-1$, ..., no more than $\delta _B^{n_1} M_1$ in period 1. Thus, $m_0\ge 1-\delta _B^{n_1}M_1$. Similarly, we have

$$\begin{aligned} m_{j-1}\ge 1-\delta _j^{n_j}M_j,\quad j=1,\ldots ,r(\omega ), \end{aligned}$$

(3)

where $\delta _j= \delta _A$ if j is even, and $\delta _j= \delta _B$ if j is odd.

If player B rejects A’s offers in period $t=1,\ldots ,n_1$, he will get no less than $m_1$ in period $n_1+1$ in any SPE. Thus, due to Lemma 6, A will offer B no less than $\delta _B^{n_1} m_1$ in period $t=1$. Hence, $M_0\le 1-\delta _B^{n_1}m_1$. Similarly, we have

$$\begin{aligned} M_{j-1}\le 1-\delta _j^{n_j}m_j,\quad j=1,\ldots ,r(\omega ), \end{aligned}$$

(4)

where $\delta _j= \delta _A$ if j is even, and $\delta _j= \delta _B$ if j is odd.

If $\omega$ is finite, by repeatedly using (3) and (4), we have

$$\begin{aligned} m_0 \ge 1-\delta _B^{n_1}M_1\ge & {} 1-\delta _B^{n_1}+\delta _B^{n_1}\delta _A^{n_2}m_2\ge 1-\delta _B^{n_1}+\delta _B^{n_1}\delta _A^{n_2}-\delta _B^{n_1+n_3}\delta _A^{n_2}M_3\\\ge & {} \cdots \ge \sum _{k=0}^{r(\omega )-1}(-1)^k p(\omega ,k)+(-1)^{r(\omega )} p(\omega ,r(\omega ))\overline{m}_{r(\omega )},\\ M_0\le 1-\delta _B^{n_1}m_1\le & {} 1-\delta _B^{n_1}+\delta _B^{n_1}\delta _A^{n_2}M_2\le 1-\delta _B^{n_1}+\delta _B^{n_1}\delta _A^{n_2}-\delta _B^{n_1+n_3}\delta _A^{n_2}m_3\\\le & {} \cdots \le \sum _{k=0}^{r(\omega )-1}(-1)^k p(\omega ,k)+(-1)^{r(\omega )} p(\omega ,r(\omega ))\underline{m}_{r(\omega )}, \end{aligned}$$

where $\overline{m}_{r(\omega )}= m_{r(\omega )}$, $\underline{m}_{r(\omega )}= M_{r(\omega )}$ if $r(\omega )$ is even, and $\overline{m}_{r(\omega )}= M_{r(\omega )}$, $\underline{m}_{r(\omega )}= m_{r(\omega )}$ if $r(\omega )$ is odd. However, in the last period of the game $t_{r(\omega )}=T(\omega )$, in any SPE, the proposer always suggests the agreement in which her own share is one. Hence $m_{r(\omega )}=M_{r(\omega )}=1$. Since $m_0\le M_0$, we have $m_0= M_0=\sum _{k=0}^{r(\omega )}(-1)^k p(\omega ,k)=\theta (\omega )$.

If $\omega$ is infinite, we can similarly get

$$\begin{aligned}&\sum _{k=0}^{j-1}(-1)^k p(\omega ,k)+(-1)^jp(\omega ,j)\overline{m}_j\\&\quad \le m_0\le M_0\le \sum _{k=0}^{j-1}(-1)^k p(\omega ,k)+(-1)^jp(\omega ,j)\underline{m}_j \end{aligned}$$

for each $j\ge 1$, where $\overline{m}_j= m_j$, $\underline{m}_j= M_j$ if j is even, and $\overline{m}_{j}= M_{j}$, $\underline{m}_{j}= m_{j}$ if j is odd. Note that $\lim _{j\rightarrow \infty }(-1)^jp(\omega ,j)\overline{m}_j=\lim _{j\rightarrow \infty }(-1)^jp(\omega ,j)\underline{m}_j=0$. Due to Lemma 1, let $j\rightarrow \infty$, we have $m_0= M_0=\sum _{k=0}^{\infty }(-1)^k p(\omega ,k)=\theta (\omega )$. Hence, we have proved the theorem. $\square$

Proof

(Proof of Theorem 2)

We shall only prove that for any $x\in [1-\delta _B,1]$, there exists $\omega \in {\varOmega }_A$, such that the payoff pair supported in SPE by $\omega$ is $(x,1-x)$. For any $x\in [0,1-\delta _B)\subset [0,\delta _A)$, a similar proof applies for $\omega \in {\varOmega }_B$.

We can construct a procedure $\omega \in {\varOmega }_A$ inductively as follows. If $x=1$, then $\omega =A$ suffices for the proof. Otherwise, we have $1-\delta _B\le x<1$. Let $n_1$ be the maximal integer, such that $1-\delta _B^{n_1}$ is not greater than x. That is

$$\begin{aligned} 1-\delta _B^{n_1}\le x<1-\delta _B^{n_1+1}. \end{aligned}$$

(5)

If the equality in (5) holds, that is, $x=1-\delta _B^{n_1}$, we can implement $(x,1-x)$ as an SPE outcome by $\omega =A^{n_1}B$. Otherwise, we have $1-\delta _B^{n_1}<x<1-\delta _B^{n_1+1}$. In this case, let $n_2$ be the maximal integer, such that $1-\delta _B^{n_1}+\delta _A^{n_2}\delta _B^{n_1}$ is not less than x. That is

$$\begin{aligned} 1-\delta _B^{n_1}+\delta _A^{n_2+1}\delta _B^{n_1}<x\le 1-\delta _B^{n_1}+\delta _A^{n_2}\delta _B^{n_1}. \end{aligned}$$

(6)

If the equality in (6) holds, that is, $x=1-\delta _B^{n_1}+\delta _A^{n_2}\delta _B^{n_1}$, we can implement $(x,1-x)$ in SPE by $\omega =A^{n_1}B^{n_2}A$. Otherwise, we have $1-\delta _B^{n_1}+\delta _A^{n_2+1}\delta _B^{n_1}<x<1-\delta _B^{n_1}+\delta _A^{n_2}\delta _B^{n_1}$, and should continue to construct $\omega$.

Now, suppose by induction that $n_1,\ldots ,n_r$ have already been defined, where $r\ge 2$. If $x=\sum _{k=0}^r(-1)^kp(\omega ,k)$, then $(x,1-x)$ can be supported in SPE by $\omega =A^{n_1}B^{n_2}\cdots A^{n_r}B$ if r is odd, or by $\omega =A^{n_1}B^{n_2}\cdots B^{n_r}A$ if r is even. Otherwise, if r is odd, then

$$\begin{aligned} \sum _{k=0}^r(-1)^kp(\omega ,k)<x<\sum _{k=0}^{r-1}(-1)^kp(\omega ,k)-\delta _Bp(\omega ,r), \end{aligned}$$

(7)

while if r is even, then

$$\begin{aligned} \sum _{k=0}^{r-1}(-1)^kp(\omega ,k)+\delta _Ap(\omega ,r)<x<\sum _{k=0}^r(-1)^kp(\omega ,k). \end{aligned}$$

In either case, we need to define $n_{r+1}$. If r is odd, let $n_{r+1}$ be the integer, such that

$$\begin{aligned} \sum _{k=0}^r(-1)^kp(\omega ,k)+\delta _Ap(\omega ,r+1)<x\le \sum _{k=0}^{r+1}(-1)^kp(\omega ,k). \end{aligned}$$

(8)

If r is even, let $n_{r+1}$ be the integer, such that

$$\begin{aligned} \sum _{k=0}^{r+1}(-1)^kp(\omega ,k)\le x<\sum _{k=0}^r(-1)^kp(\omega ,k)-\delta _Bp(\omega ,r+1). \end{aligned}$$

(9)

We need to show that $n_{r+1}$ is well defined, that is, there exists exactly one integer $n_{r+1}$, such that (8) or (9) holds. If r is odd, $\sum _{k=0}^{r+1}(-1)^kp(\omega ,k)$ is decreasing in $n_{r+1}$. When $n_{r+1}=1$, according to (7), we have

$$\begin{aligned} \begin{aligned} \sum _{k=0}^{r+1}(-1)^kp(\omega ,k)&=\sum _{k=0}^{r}(-1)^kp(\omega ,k)+\delta _Ap(\omega ,r)\\&\ge \sum _{k=0}^{r}(-1)^kp(\omega ,k)+(1-\delta _B)p(\omega ,r)\\&=\sum _{k=0}^{r-1}(-1)^kp(\omega ,k)-\delta _Bp(\omega ,r)>x, \end{aligned} \end{aligned}$$

(10)

while as $n_{r+1}\rightarrow \infty$, $\sum _{k=0}^{r+1}(-1)^kp(\omega ,k)\rightarrow \sum _{k=0}^{r}(-1)^kp(\omega ,k)<x$. Thus, there exists a unique integer $n_{r+1}$ that satisfies (8), that is, $n_{r+1}$ is well defined. Similarly, we can prove $n_{r+1}$ is well defined if r is even. We have thus finished defining $n_{r+1}$.

Given $x\in [1-\delta _B,1]$, if there exists an integer h, such that for $n_1,\ldots ,n_h$ defined above, $x=\sum _{k=0}^{h}(-1)^kp(\omega ,k)$, then from Theorem 1, the payoff pair $(x,1-x)$ can be supported as an SPE outcome by $\omega =A^{n_1}B^{n_2}\cdots A^{n_h}B$ if h is odd, or by $\omega =A^{n_1}B^{n_2}\cdots B^{n_h}A$ if h is even. If such a finite sequence of integers cannot be found, we will eventually get an infinite sequence $n_1,n_2,\ldots$ and a corresponding infinite procedure $\omega =A^{n_1}B^{n_2}\cdots$. Let

$$\begin{aligned} z_r=\sum _{k=0}^{r}(-1)^kp(\omega ,k),\quad r=1,2,\ldots . \end{aligned}$$

The elements in the sequence $\{z_1,z_2,\ldots \}$ are alternately larger and smaller than x. According to Lemma 1, this sequence converges, and thus, its limit is x. That is, $x=\sum _{k=0}^{\infty }(-1)^kp(\omega ,k)$. It follows from Theorem 1 that $(x,1-x)$ can be supported as an SPE outcome by $\omega$. Thus, we have proved the theorem. Note that the condition $\delta _A+\delta _B\ge 1$ is used in the middle inequality of (10). $\square$

Proof

(Proof of Theorem 3)

The proof of the theorem proceeds in steps.

Step 1: We shall show $(\delta _A,1-\delta _B)\cap \varGamma (\delta _A,\delta _B)=\emptyset$, that is, if $\delta _A<x<1-\delta _B$, then there does not exist $\omega \in \varOmega$, such that $(x,1-x)$ can be supported as an SPE outcome by $\omega$. Suppose, on the contrary, that we can find such a procedure $\omega$, and assume without loss of generality that $\omega \in \varOmega _A$. According to Theorem 1 and Lemma 3, $x=\theta (\omega )=\sum _{k=0}^{r(\omega )}(-1)^kp(\omega ,k) \ge 1-\delta _B^{n_1}\ge 1-\delta _B$, which contradicts $x<1-\delta _{B}$. Thus, $(\delta _A,1-\delta _B)\cap \varGamma (\delta _A,\delta _B)=\emptyset$. Therefore, $m[\varGamma (\delta _A,\delta _B)]\le 1-m[(\delta _A,1-\delta _B)]=\delta _A+\delta _B$.

Step 2: We shall first prove that $(1-\delta _B+\delta _B\delta _A,1-\delta _B^2)\cap \varGamma (\delta _A,\delta _B)=\emptyset$. Suppose, on the contrary, that we can find $x\in (1-\delta _B+\delta _B\delta _A,1-\delta _B^2)$ and $\omega \in \varOmega$, such that $(x,1-x)$ can be supported as an SPE outcome by $\omega$. Since $x>1-\delta _B$, we have $\omega \in \varOmega _A$, that is, $\omega _1=1$. Again, due to Theorem 1 and Lemma 3, if $\omega _2=1$, then $x\ge 1-\delta _B^{n_1}\ge 1-\delta _B^{2}$; if $\omega _2=2$, then $x\le 1-\delta _B+\delta _B\delta _A^{n_2}\le 1-\delta _B+\delta _B\delta _A$. This contradicts $x\in (1-\delta _B+\delta _B\delta _A,1-\delta _B^2)$. Thus, $(1-\delta _B+\delta _B\delta _A,1-\delta _B^2)\cap \varGamma (\delta _A,\delta _B)=\emptyset$. Similarly, we can prove $(\delta _A^2,\delta _A-\delta _B\delta _A)\cap \varGamma (\delta _A,\delta _B)=\emptyset$. Hence, $m[\varGamma (\delta _A,\delta _B)]\le \delta _A+\delta _B-m[(1-\delta _B+\delta _B\delta _A,1-\delta _B^2)]-m[(\delta _A^2,\delta _A-\delta _B\delta _A)]=(\delta _A+\delta _B)^2$.

Assume by induction that by the first $k\ge 1$ steps, we have deleted the following $2^{k}-1$ disjoint intervals from $\varGamma (\delta _A,\delta _B)$: at step one, we delete $A^1_1=(\delta _A,1-\delta _B)$, at step two, we delete $A^2_1=(\delta _A^2,\delta _A-\delta _B\delta _A)$ and $A^2_2=(1-\delta _B+\delta _B\delta _A,1-\delta _B^2)$, $\cdots$, at step k, we delete $A^k_1=(\delta _A^k,\delta _A^{k-1}(1-\delta _B))$, $\ldots$, $A^k_{2^{k-1}}=(1-\delta _B^{k-1}(1-\delta _A),1-\delta _B^k)$. The exact structure of each $A^k_i$ can be derived according to the inductive algorithm described at step $k+1$ below. There remain $2^{k}$ intervals, whose total measure is $1-\sum _{s=1}^k\sum _{i=1}^{2^{k-1}}m[A^s_i]=(\delta _A+\delta _B)^k$. Thus, $m[\varGamma (\delta _A,\delta _B)]\le (\delta _A+\delta _B)^k$.

Now, at step $k+1$, we further delete $2^{k}$ disjoint intervals $A^{k+1}_i$, $i=1,\ldots ,2^{k}$, from $\varGamma (\delta _A,\delta _B)$. More specifically, in each remaining interval $[P_i,Q_i]$ after step k, we delete $A^{k+1}_i=\big (P_i+\delta _A(Q_i-P_i),Q_i-\delta _B(Q_i-P_i)\big )$. There remain two disjoint intervals in $[P_i,Q_i]$: $[P_i,P_i+\delta _A(Q_i-P_i)]$ and $[Q_i-\delta _B(Q_i-P_i),Q_i]$, whose total measure is $(\delta _A+\delta _B)(Q_i-P_i)$. Therefore, after step $k+1$, there remain $2^{k+1}$ intervals, whose total measure is $(\delta _A+\delta _B)^{k+1}$.

To show $m[\varGamma (\delta _A,\delta _B)]\le (\delta _A+\delta _B)^{k+1}$, we still have to prove that $A^{k+1}_i\cap \varGamma (\delta _A,\delta _B)=\emptyset$, $i=1,\ldots ,2^{k}$. Without loss of generality, we only consider $A^{k+1}_1=\big (\delta _A^{k+1},\delta _A^{k}(1-\delta _B)\big )\subset [P_1,Q_1]=[0,\delta _A^k]$. Suppose, on the contrary, that we can find $x\in A^{k+1}_1$ and $\omega \in \varOmega$, such that $(x,1-x)$ can be supported as an SPE outcome by $\omega$. Since $0<x<\delta _A<1-\delta _B$, we may write $\omega =B^{n_1}A^{n_2}\cdots$, then $x=\delta _A^{n_1}-\delta _A^{n_1}\delta _B^{n_2}+\cdots$. If $n_1\ge k+1$, then $x<\delta _A^{n_1}\le \delta _A^{k+1}$, which contradicts $x> \delta _A^{k+1}$. If $n_1\le k$, then $x \ge \delta _A^{n_1}(1-\delta _B^{n_2})\ge \delta _A^{k}(1-\delta _B)$, which contradicts $x<\delta _A^{k}(1-\delta _B)$. Thus, $A^{k+1}_1\cap \varGamma (\delta _A,\delta _B)=\emptyset$.

Hence, we have proved $m[\varGamma (\delta _A,\delta _B)]\le (\delta _A+\delta _B)^k$, $\forall k\ge 1$. Since $\delta _A+\delta _B<1$, we have $(\delta _A+\delta _B)^k\rightarrow 0$ as $k\rightarrow \infty$. Thus, $m[\varGamma (\delta _A,\delta _B)]=0$.

Proof

(Proof of Corollary 1)

In the proof of Theorem 3, we have shown that the set $[0,1]\backslash \varGamma (\delta _A,\delta _B)$ can be constructed as the union of infinitely many disjoint intervals: $A^1_1=(\delta _A,1-\delta _B)$, $A^2_1=(\delta _A^2,\delta _A-\delta _B\delta _A)$, $A^2_2=(1-\delta _B+\delta _B\delta _A,1-\delta _B^2)$,..., $A^k_1=(\delta _A^k,\delta _A^{k-1}(1-\delta _B))$, $\ldots$, $A^k_{2^{k-1}}=(1-\delta _B^{k-1}(1-\delta _A),1-\delta _B^k)$,...,. Let $\{C_j\}_{j=1,2,\ldots }$ be a sequence, such that (a) for each $A^k_i$, $i=1,\ldots ,2^{k-1}$, $k=1,2,\ldots$, there is one and only one $C_j=A^k_i$, and (b) $m(C_j)\ge m(C_{j+1})$, $j=1,2,\ldots$,. Since $\varGamma ^\lambda (\delta _A,\delta _B)=\varGamma ^\lambda (\delta _A,\delta _B)\cap [0,1] = \varGamma ^\lambda (\delta _A,\delta _B) \cap \left( \cup _{j=1}^\infty C_j \cup \varGamma (\delta _A,\delta _B)\right) = \cup _{j=1}^\infty \left( \varGamma ^\lambda (\delta _A,\delta _B)\cap C_j\right) \cup \varGamma (\delta _A,\delta _B)$, due to Theorem 3, we have

$$\begin{aligned} m[\varGamma ^\lambda (\delta _A,\delta _B)]=\sum _{j=1}^\infty m[C_j\cap \varGamma ^\lambda (\delta _A,\delta _B)]. \end{aligned}$$

(11)

Since $\delta _A\in \varGamma (\delta _A,\delta _B)$, $1-\delta _B\in \varGamma (\delta _A,\delta _B)$ but $A^1_1\cap \varGamma (\delta _A,\delta _B)=\emptyset$, we have $A^1_1\cap \varGamma ^\lambda (\delta _A,\delta _B)= (\delta _A,\delta _A+\lambda ]\cup [1-\delta _B-\lambda ,1-\delta _B)$, and thus, $m\big [C_1\cap \varGamma ^\lambda (\delta _A,\delta _B)\big ]= m\big [A^1_1\cap \varGamma ^\lambda (\delta _A,\delta _B)\big ]=\min \{2\lambda ,1-\delta _A-\delta _B\} = \min \{2\lambda ,m(C_1)\}$. Similarly, for each $C_j$, $j=1,2,\ldots$, we have $m\big [C_j\cap \varGamma ^\lambda (\delta _A,\delta _B)\big ]=\min \{2\lambda ,m(C_j)\}$.

For each $\varepsilon$, let $\mu$ be a real number, such that $\sum _{j=1}^{[{\varepsilon }/{(4\mu )}]}m(C_j) >1-\frac{\varepsilon }{2}$, where $[{\varepsilon }/{(4\mu )}]$ is the integer part of ${\varepsilon }/{(4\mu )}$. Due to the proof of Theorem 3, such $\mu$ can be found when $\mu$ is small enough. When $\lambda <\mu$, according to (11), we have

$$\begin{aligned} \begin{aligned} m[\varGamma ^\lambda (\delta _A,\delta _B)]&= \sum _{j=1}^{[{\varepsilon }/{(4\lambda )}]} m[C_j\cap \varGamma ^\lambda (\delta _A,\delta _B)] + \sum _{j=[{\varepsilon }/{(4\lambda )}]+1}^{\infty } m[C_j\cap \varGamma ^\lambda (\delta _A,\delta _B)]\\&\le \frac{\varepsilon }{4\lambda }\cdot (2\lambda ) + \sum _{j=[{\varepsilon }/{(4\lambda )}]+1}^{\infty } m(C_j)\\&=\frac{\varepsilon }{2} + 1-\sum _{j=1}^{[{\varepsilon }/{(4\lambda )}]}m(C_j)<\frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon . \end{aligned} \end{aligned}$$

This is the conclusion we want to prove. $\square$