The Complex Fuzzy Measure

Abstract

In this paper, we define the concept of complex Fuzzy measure, which is different from the concept of complex Fuzzy measure in [2], and discuss its properties and theorems. On the basis of the concept of complex Fuzzy measurable function in [2], we study its convergence theorem. It builds the certain foundation for the research of complex Fuzzy integral.

1 Introduction

In 1990–1991, Buckley [1] proposed the concept of fuzzy complex numbers and fuzzy complex-valued function. In 1997, Qiu Jiqing [2–5] firstly proposed the concept of the complex fuzzy measure on the basis of classical measure theory method. Since 2000, According to this issue, Ma shengquan [6] has done some exploratory work, and made a series of achievement in this field. The theory of fuzzy complex valued measure is an important part of fuzzy complex analysis, which has a strong background of practical application [7]. For instance it can use in the fuzzy system identification, fuzzy control, multi-classifier system design and other fields. The development of theoretical research of fuzzy complex valued measure is slow, because it is much more complicated than Fuzzy real-valued measure. The complex Fuzzy measure which defined in this paper is different from that in paper [2], the concept of Fuzzy measure was redefined, which distinguished between the real and imaginary parts in order to facilitate research.

2 Complex Fuzzy Measure

$\hat{{R}}^{+}$ denote positive real set, $\hat{{C}}^{+}$ denote the set of complex number on $\hat{{R}}^{+}$ [8].

Definition 2.1

Let $X$ be a nonempty set, $\text {F}$ be a $\sigma -$ algebra, comprising of the subset of $X ,$ the mapping $\mu \text {:F} \rightarrow {\hat{{\mathrm{C}}}}^{+}$ is set function, satisfying:

• (1) $\mu (\emptyset )=0$;

• (2) (monotonicity) If $A,B \in \text {F}$ and $A\subseteq B$ ,then $Re(\mu (A)) \le Re(\mu (B))$ and $Im(\mu (A))\le Im(\mu (B)).$ Denote $\mu (A)\le \mu (B)$

• (3) if $A_n \in \text {F} (n=1,2,\cdots )$, $A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n \subseteq \cdots$ then

  $$\begin{aligned} \mu (\bigcup _{n=1}^\infty {A_n } )=\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A_n ) \end{aligned}$$

• (4) if $A_n \in \text {F} (n=1,2,\ldots )$, $A_1 \supseteq A_2 \supseteq \cdots \supseteq A_n \supseteq \cdots$, and $\exists n_0$ such that $\text {Re}(\mu (A_{n_0 } ))<\infty ,\text {Im}(\mu (A_{n_0 } ))<\infty$ then $\mu ({\mathop \bigcap \limits _{n=1}^\infty {A_n}})=\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A_n )$. Then $\mu$ is called as complex Fuzzy measure on $\text {F}, (\text {X,F,}\mu )$ is called as complex Fuzzy measure space.

Definition 2.2

Fuzzy complex measure $\mu$ is said to be zero-additive, if for arbitrary $E,F\in \text {F}, \mu (F)=0$ and $E\cap F=\varphi$, then $\mu (E\cup F)=\mu (E)$.

Theorem 2.1

(X, F, $\mu$) is complex Fuzzy measure space, The following propositions are equivalence.

• (1) $\mu$ is zero-additive ;

• (2) since $\mu (F)=0$, then for arbitrary $E,F\in \text {F}$, such that $\mu (E\cup F)=\mu (E)$;

• (3) since $\mu (F)=0$, then for arbitrary $E,F\in \text {F}$, such that $\mu (E\backslash F)=\mu (E)$:

Proof.

(1)$\Rightarrow$(2):

$E\cup F=E\cup (F\backslash E),$ since $\mu$ is nonnegative monotony, then for $\mu (F)=0$, such that

$\mu (F\backslash E)\le \mu (F)=0,$ therefore $\mu (F\backslash E)=0$. Applying the zero-additive of $\mu ,\mu (F\backslash E)=0$ and $E\cap (F\backslash E)=\varphi$, then

$$\begin{aligned} \mu (E\cup F)=\mu (E\cup (F\backslash E))=\mu (E). \end{aligned}$$

(2)$\Rightarrow$(3):

Due to $E=(E\backslash F)\cup (E\cap F)$, and $\mu (F)=0$, then $\mu (E\cap F)=0$.

We know $\mu (E)=\mu ((E\backslash F)\cup (E\cap F))=\mu (E\backslash F)$ from the proposition (2).(3)$\Rightarrow$(1):

Due to $E\cap F=\varphi$, then $E=(E\cup F)\backslash F$.

If $\mu (F)=0$ and $E\cap F=\varphi$, we can know

$$\begin{aligned} \mu (E)=\mu ((E\cup F)\backslash F)=\mu (E\cup F) \end{aligned}$$

from the proposition (3).

Theorem 2.2

Suppose $\mu$ is complex Fuzzy measure of zero-additive, $A\in \text {F}$, there is a descending sequence of $\{B_n \}\subset \text {F}$, $(B_1 \supseteq B_2 \supseteq \cdots )$, if $\mu (B_n )\rightarrow 0$, then

• (1) $\mu (A\backslash B_n )\rightarrow \mu (A)$;

• (2) whereupon $\text {Re}(\mu (A))<\infty ,$ and $\text {Im}(\mu (A))<\infty$ , and if exists $\text {Re}(\mu (A\cup B_{n_0 } ))<\infty$ and $\text {Im}(\mu (A\cup B_{n_0 } ))<\infty$ , therefore $\mu (A\cup B_n )\rightarrow \mu (A)$.

Proof.

(1) since $\{A\backslash B_n \}$ is ascending sequence and if $\mu$ is lower-continuous, we can know $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\backslash B_n )=\mu ({\mathop \bigcup \limits _{n=1}^\infty {(A\backslash B_n )}})=\mu (A\backslash {\mathop \bigcap \limits _{n=1}^\infty {B_n)}}$.

Applying $\mu$ is upper-continuous, $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (B_n )=\mu ({\mathop \bigcap \limits _{n=1}^\infty {B_n}})=0$, since $\mu$ is zero-additive, from the Theorem 1, then $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\backslash B_n )=\mu (A)$.

(2) $\{A\cup B_n \}$ is descending sequence, $Re(\mu (A))<\infty ,$ and $Im(\mu (A))<\infty$, and exist $Re(\mu (A\cup B_{n_0 } ))<\infty$ and $Im(\mu (A\cup B_{n_0 } ))<\infty$, due to $\mu$ is upper-continuous,

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\cup B_n )=\mu (\bigcap _{n=1}^\infty {(A\cup B_n } ))=\mu (A\cup (\bigcap _{n=1}^\infty {B_n } )). \end{aligned}$$

Since $\mu$ is zero-additive, and $\mu ({\mathop \bigcap \limits _{n=1}^\infty {B_n}})=0$, so we know

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\cup B_n )=\mu (A) \end{aligned}$$

according to Theorem 1.

Definition 2.3

Let $\text {(X,F)}$ be measurable space, the mapping $\mu \text {:F}\rightarrow {\hat{{\mathrm{C}}}}^{+}$ is set function, Complex fuzzy measure $\mu$ is upper-self-continuous, if for arbitrary $A,B_n \in \text {F}$, and $A\cap B_n =\Phi$, $\mathop {\lim }\limits _{n\rightarrow \infty } B_n =0$, then

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\backslash B_n )=\mu (A). \end{aligned}$$

Complex fuzzy measure $\mu$ is lower-self-continuous, if for arbitrary $A,B_n \in \text {F}$ and $B_n \subseteq A$, $\mathop {\lim }\limits _{n\rightarrow \infty } B_n =0$, then $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\backslash B_n )=\mu (A)$.

Complex fuzzy measure $\mu$ is self-continuous, if and only if $\mu$ is not only upper-self-continuous but also lower-self-continuous.

Theorem 2.3

Suppose $\mu$ is complex Fuzzy measure on $\text {(X,F)}$, then

• (1) $\mu$ is upper-self-continuous if and only if $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\cup B_n )=\mu (A)$ if for arbitrary $A,B_n \in \text {F}$ and $\mathop {\lim }\limits _{n\rightarrow \infty } B_n =0$.

• (2) $\mu$ is lower-self-continuous if and only if $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\backslash B_n )=\mu (A)$ if for arbitrary $A,B_n \in \text {F}$ and $\mathop {\lim }\limits _{n\rightarrow \infty } B_n =0$.

Proof.

The necessity is easily proved from the definition.2.3

• (1) Sufficiency: Let $E_n =B_n \backslash A$, we know $\mu (E_n )\le \mu (B_n )$, therefore $E_n \cap A=\varphi$, and $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (E_n )=0$, then $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\cup E_n )=\mu (A)$, so $\mu$ is upper-self-continuous.

• (2) Sufficiency: Let $E_n =B_n \cap A$, we know $\mu (E_n )\le \mu (B_n )$, therefore $E_n \subseteq A$, and $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (E_n )=0$, then $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (A\backslash E_n )=\mu (A)$, so $\mu$ is lower-self-continuous

Definition 2.4

Let $\text {(X, F)}$ be measurable space, the mapping $\mu \text {:F}\rightarrow {\hat{{\mathrm{C}}}}^{+}$ is set function,

• (1) Suppose for arbitrary $\varepsilon _i >0$, exists $\delta _i =\delta (\varepsilon _i )>0(i=1,2)$, where $\varepsilon =\varepsilon _1 +i\varepsilon _2 , \delta =\delta _1 +i\delta _2$, Complex fuzzy measure $\mu$ is uniform-upper-self-continuous, if for arbitrary $A,B\in \text {F}$ and $\mu (B)\le \delta$ , then $\mu (A\cup B)\le \mu (A)+\varepsilon$.

• (2) Suppose for arbitrary $\varepsilon _i >0$, exists $\delta _i =\delta (\varepsilon _i )>0(i=1,2)$, where $\varepsilon =\varepsilon _1 +i\varepsilon _2$, $\delta =\delta _1 +i\delta _2$, Complex fuzzy measure $\mu$ is uniform-lower-self- continuous, if for arbitrary $A,B\in \text {F}$ and $\mu (B)\le \delta$, then $\mu (A)-\varepsilon \le \mu (A\backslash B)$.

• (3) Complex fuzzy measure $\mu$ is uniform-self-continuous, if and only if $\mu$ is not only uniform-upper-self-continuous but also uniform-lower-self-continuous.

Theorem 2.4

Suppose set function $\mu$ is uniform-upper-self-continuous (uniform-lower-self-continuous), then $\mu$ is upper-self-continuous (lower-self- continuous).

Proof.

It is obvious.

Theorem 2.5

Suppose $\mu$ is complex Fuzzy measure on $\text {(X,F)}$, The following propositions are equivalence.

• (1) $\mu$ is uniform-self-continuous;

• (2) $\mu$ is uniform-upper-self-continuous;

• (3) $\mu$ is uniform-lower-self-continuous.

Proof.

$(1)\Rightarrow (2)$ It is obvious.

$(2)\Rightarrow (3)$: Since $\mu$ is uniform-upper-self-continuous, so if for arbitrary $\varepsilon _i >0,\exists \delta _i =\delta (\varepsilon _i )>0(i=1,2)$, where $\varepsilon =\varepsilon _1 +i\varepsilon _2 ,\delta =\delta _1 +i\delta _2$, and for arbitrary ${A}',{B}'\in \text {F},\mu ({B}')\le \delta ,$then $\mu ({A}')-\varepsilon \le \mu ({A}'\cup {B}')\le \mu ({A}')+\varepsilon$. if for arbitrary $A,B\in \text {F},\mu (B)\le \delta ,$ Let${A}'=A\backslash B,{B}'=A\cap B,\mu ({B}')\le \mu (B)\le \delta ,$

if $\mu (A\backslash B)-\varepsilon \le \mu ({A}'\cup {B}')\le \mu (A\backslash B)+\varepsilon$, then $\mu (A)-\varepsilon \le \mu (A\backslash B)\le \mu (A)+\varepsilon$. It means $\mu$ is uniform-lower-self-continuous.

$(3)\Rightarrow (1)$: Since $\mu$ is uniform-lower-self-continuous, if for arbitrary

$\varepsilon _i >0,\exists \delta _i =\delta (\varepsilon _i )>0(i=1,2),$ where $\varepsilon =\varepsilon _1 +i\varepsilon _2 ,\delta =\delta _1 +i\delta _2$, and for arbitrary ${A}',{B}'\in \text {F},\mu ({B}')\le \delta ,$ then $\mu ({A}')-\varepsilon \le \mu ({A}'\backslash {B}')\le \mu ({A}')+\varepsilon$. if for arbitrary

$A,B\in \text {F}, \mu (B)\le \delta ,$ Let${A}'=A\cup B,{B}'=A\cap B,\mu ({B}')\le \mu (B)\le \delta ,$

therefore

$$\begin{aligned} \mu ({A^\prime }\backslash {B\prime })\ge \mu ({A^\prime })-\varepsilon \ge \mu (A)-\varepsilon . \end{aligned}$$

Again let ${{A}^{\prime \prime }}=(A\cup B)\backslash (A\cap B),{{B}^{\prime \prime }}=B\backslash A$, then ${{A}^{\prime \prime }}\backslash {{B}^{\prime \prime }}=A\backslash B$, and $\mu ({{B}^{\prime \prime }})\le \mu (B)\le \delta ,$ so

$$\begin{aligned} \mu ({{A}^{\prime \prime }})-\varepsilon \le \mu ({{A}^{\prime \prime }}\backslash {{B}^{\prime \prime }})=\mu (A\backslash B) \Rightarrow \mu (A\backslash B)+\varepsilon \le \mu (A)+\varepsilon . \end{aligned}$$

Again let $A=E,B=F\backslash E$, then $(A\cup B)\backslash (A\cap B)=E\cup F , \mu (B)\le \mu (F)\le \delta ,$ so $\mu (E)-\varepsilon \le \mu (E\cup F)\le \mu (E)+\varepsilon$. It means $\mu$ is uniform-upper-self-continuous.

So $\mu$ is uniform-self-continuous.

Definition 2.5

Let $\text {(X,F)}$ be measurable space, the mapping $\mu \text {:F}\rightarrow {\hat{{\mathrm{C}}}}^{+}$ is set function, If for arbitrary $\{B_n \}\subseteq \text {A}, B_1 \supseteq B_2 \supseteq \cdots$, if $\exists n_0 ,\forall n>n_0 , Re(\mu (B_n ))<\infty$, $Im(\mu (B_n ))<\infty$ and ${\mathop \bigcap \limits _{n=1}^\infty {B_n}} =\varphi$, there must be $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (B_n )=0$, so $\mu$ is called zero-upper-continuous.

Theorem 2.6

$\mu$ is nonnegative monotonic ascending set function, and zero-upper-continuous, Then if $\mu$ is upper-self- continuous, then $\mu$ is upper- continuous;

if $\mu$ is limit and lower-self- continuous, then $\mu$ is lower- continuous.

Proof

(1) If $\{A_n \}\subseteq \text {F}, A_1 \supseteq A_2 \supseteq \cdots$, exist $Re(\mu (B_n ))<\infty ,Im(\mu (B_n ))<\infty$, let

$A={\mathop \bigcap \limits _{n=1}^\infty {A_n}} ,B_n =A_n \backslash A \quad (n=1,2,\cdots ),$then $B_1 \supseteq B_2 \supseteq \cdots$, and ${\mathop \bigcap \limits _{n=1}^\infty {B_n}} =\varphi$.if for

arbitrary

$$\begin{aligned} n>n_0 ,Re(\mu (B{ }_n))\le Re(\mu (B_{n_0 } ))\le Re(\mu (A_{n_0 } ))<\infty , \end{aligned}$$

$\text {Im}(\mu (B{ }_n))\le \text {Im}(\mu (B_{n_0 } ))\le \text {Im}(\mu (A_{n_0 } ))<\infty .$ Since $\mu$ is zero-upper-continuous, we know $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (B_n )=0$. Due to $A_n =A\cup B_n ,A\cap B_n =\varphi$, and $\mu$ is upper-self-continuous, if $\mu (A_n )=\mu (A\cup B_n )$, then $\mu (A)=\mu ({\mathop \bigcap \limits _{n=1}^\infty {A_n}})$.So $\mu$ is upper- continuous.

(2)The proof is similar to (1) above.

3 Complex Fuzzy Measurable Function

$R$ denote real set, $C$ denote the set of complex number on $R$.

Definition 3.1

[2] Suppose $(\text {X,F,}\mu )$ is complex Fuzzy measure space, the mapping $\tilde{f}:X\rightarrow C$ is called complex Fuzzy measurable function, if for arbitrary $a+bi\in C$, then $\{x\in X\left| {Re[\tilde{f}(x)]\ge a,Im[\tilde{f}(x)]\ge b\}\in } \right. \text {F}$.

Definition 3.2

Suppose $(\text {X,F,}\mu )$ is complex Fuzzy measure space, $\tilde{f}_n (n=1,2,\cdots ),\tilde{f}$ is complex fuzzy measurable function, for arbitrary $A\in \text {F}$,

• (1) $\{\tilde{f}_n \}$ almost everywhere converge to $\tilde{f}$ on $A$, denote $\tilde{f}_n {\mathop {\rightarrow }\limits ^{a.e.}}\tilde{f}$, if there exists $B\in \text {F}$, such that $\mu (B)=0$ , then $\{\tilde{f}_n \}$ with pointwise convergence to $\tilde{f}$ on $A\backslash B$.

• (2) $\{\tilde{f}_n \}$ pseudo-almost everywhere converge to $\tilde{f}$ on $A$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.}\tilde{f}$, if there exists $B\in \text {F}$, such that $\mu (A\backslash B)=\mu (A)$, then $\{\tilde{f}_n \}$ with pointwise convergence to $\tilde{f}$ on $A\backslash B$.

• (3) $\{\tilde{f}_n \}$ almost everywhere uniformly converge to $\tilde{f}$ on $A$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.u.}\tilde{f}$, if there exists $B\in \text {F}$, such that $\mu (B)=0$, then $\{\tilde{f}_n \}$ with pointwise uniform convergence to $\tilde{f}$ on $A\backslash B$.

• (4) $\{\tilde{f}_n \}$ pseudo-almost everywhere uniformly converge to $\tilde{f}$ on $A$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.u.}\tilde{f}$, if there exists $B\in$ F, such that $\mu (A\backslash B)=\mu (A)$, then $\{\tilde{f}_n \}$ with pointwise uniform convergence to $\tilde{f}$ on $A\backslash B$.

• (5) $\{\tilde{f}_n \}$ almost uniformly converge to $\tilde{f}$ on $A$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.u.}\tilde{f}$, if there exists set sequence$\{E_k \}$ on $\text {F}$, such that $\tilde{\mu }(E_k )\rightarrow 0$ ,and for arbitrary $k$, then $\{\tilde{f}_n \}$ with pointwise uniform convergence to $\tilde{f}$ on $A\backslash E_k$.

• (6) $\{\tilde{f}_n \}$ pseudo-almost uniformly converge to $\tilde{f}$ on $A$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.u.}\tilde{f}$, if there exists set sequence $\{E_k \}$ on $\text {F}$, such that $\mu (A\backslash E_k )\rightarrow \mu (A)$, and for arbitrary $k$, then $\{\tilde{f}_n \}$ with pointwise uniform convergence to $\tilde{f}$ on $A\backslash E_k$.

• (7) $\{\tilde{f}_n \}$ converge in complex fuzzy measure $\mu$ to $\tilde{f}$ on $A$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{u.}\tilde{f}$, if for arbitrary $\varepsilon =\varepsilon _1 +i\varepsilon _2 ,\varepsilon _1 ,\varepsilon _2 >0$, such that

  $$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \mu (\{x\left| {Re\left| {\tilde{f}_n -\tilde{f}} \right| } \right. \ge \varepsilon _1 ,Im\left| {\tilde{f}_n -\tilde{f}} \right| \ge \varepsilon _2 \}\cap A)=0. \end{aligned}$$

• (8) $\{\tilde{f}_n \}$ converge in pseudo complex fuzzy measure $\tilde{\mu }$ to $\tilde{f}$ on $A$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.u.}\tilde{f}$, if for arbitrary $\varepsilon >0$, such that

  $$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \mu (\{x\left| {Re\left| {\tilde{f}_n -\tilde{f}} \right| } \right. <\varepsilon _1 ,Im\left| {\tilde{f}_n -\tilde{f}} \right| <\varepsilon _2 \}\cap A)=\mu (A). \end{aligned}$$

Theorem 3.1

Suppose $(\text {X,F,}\mu )$ is complex Fuzzy measure space, $\tilde{f}_n (n=1,2,\cdots ),\tilde{f}$ is complex fuzzy measurable function, for arbitrary $A\in$ F,

• (1) $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.}\tilde{f}$ if and only if $\{\tilde{f}_n \}$ converge to $\tilde{f}$ on $A\backslash E_k$, if there exists set sequence $\{E_k \}$ on F, such that $\mu (E_k )\rightarrow 0$, and for arbitrary $k$.

• (2) $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.}\tilde{f}$ if and only if $\{\tilde{f}_n \}$ converge to $\tilde{f}$ on $A\backslash E_k$, if there exists set sequence $\{E_k \}$ on F, such that $\mu (A\backslash E_k )\rightarrow \mu (A)$, and for arbitrary $k$.

• (3) $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.u.}\tilde{f}$ if and only if $\left| {Re(\tilde{f}-\tilde{f}_k )} \right| <\varepsilon _1$ and $\left| {Im(\tilde{f}-\tilde{f}_k )} \right| <\varepsilon _2$, if there exists set sequence $\{E_k \}$ on F, such that $\mu (E_k )\rightarrow 0$, and for arbitrary $\varepsilon _i >0,(i=1,2)$, where $\varepsilon =\varepsilon _1 +i\varepsilon _2$, $\exists n_0,\forall n>n_0,\forall k,\forall x\in A\backslash E_k$.

• (4) $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.u.}\tilde{f}$ if and only if $\left| {Re(\tilde{f}-\tilde{f}_k )} \right| <\varepsilon _1$ and $\left| {Im(\tilde{f}-\tilde{f}_k )} \right| <\varepsilon _2$, if there exists set sequence $\{E_k \}$ on $\text {F}$, such that $\mu (A\backslash E_k )\rightarrow \mu (A)$, and for arbitrary $\varepsilon _i >0,(i=1,2)$, where $\varepsilon =\varepsilon _1 +i\varepsilon _2$, $\exists n_0, \forall n>n_0, \forall k,\forall x\in A\backslash E_k$.

Proof.

(1) If $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.}\tilde{f}$, then exists $B\in$F, $\mu (B)=0$, such that $\tilde{f}_n \rightarrow \tilde{f}$ on $A\backslash B$. Let $E_k =B (k=1,2,\cdots )$, we know $\mu (E_k )\rightarrow 0$, and $\{\tilde{f}_n \}$ converge to $\tilde{f}$ on $A\backslash E_k$, for arbitrary $k$.

Otherwise, if there exists $\{E_k \}\subseteq$F,$\mu (E_k )\rightarrow 0$, such that $\tilde{f}_n \rightarrow \tilde{f}$ on $A\backslash E_k$. Let $B_k ={\mathop \bigcap \limits _{i=1}^k {E_i}} ,B={\mathop \bigcap \limits _{k=1}^\infty {B_k}} ={\mathop \bigcap \limits _{k=1}^\infty {E_k}}$, so $\mu (B_k )\le \mu (E_k )$ and $B_1 \supseteq B_2 \supseteq \cdots$. Due to $\mu (E_k )\rightarrow 0$, there exists

$\text {Re}(\mu (B_{k_0 } ))\le \text {Re}(\mu (E_{k_0 } ))<\infty ,$and $\text {Im}(\mu (B_{k_0 } ))\le \text {Im}(\mu (E_{k_0 } ))<\infty$.

Applying the upper-continuity of $\mu$, we know that $\mu (B)=\mathop {\lim }\limits _{k\rightarrow 0} \mu (B_k )=0$. If for arbitrary $x\in A\backslash B={\mathop \bigcup \limits _{k=1}^\infty {(A\backslash E_k )}} ,\exists k_0 ,x\in A\backslash E_{k_0 }$, then $\tilde{f}_n \rightarrow \tilde{f}$, therefore $\{\tilde{f}_n \}$ converge to $\tilde{f}$ on $A\backslash B$, denote $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.}\tilde{f}$.

The proof of (2),(3),(4) is similar to (1), we omit here.

Inference 3.1 If $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.u.}\tilde{f}$, then $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.u.}\tilde{f}$;If $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.u.}\tilde{f}$, then $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.u.}\tilde{f}$.

Theorem 3.2

Suppose complex Fuzzy measure $\mu$ is lower-self- continuity, for $A\in$F,

If $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.}\tilde{f}$ , then $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.}\tilde{f}$.

If $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.u.}\tilde{f}$ , then $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.u.}\tilde{f}$.

If $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.u.}\tilde{f}$ then $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.u.}\tilde{f}$.

Proof.

(1) If $\tilde{f}_n \mathop {\rightarrow }\limits ^{a.e.}\tilde{f}$ on $\text {A}$, then there exists $B\in \text {F}, \mu (B)=0$, such that $\{\tilde{f}_n \}$ converge to $\tilde{f}$ on $A\backslash B$, therefore $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.a.e.}\tilde{f}$ on $A$.

The proof of (2),(3) is similar to (1), we omit here.

Theorem 3.3

Suppose for arbitrary $A\in$F, and complex fuzzy measurable function $\tilde{f}$ and $\tilde{f}_n (n=1,2,\cdots )$, if $\tilde{f}_n \mathop {\rightarrow }\limits ^{u.}\tilde{f}$, then $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.u.}\tilde{f}$ if and only if complex fuzzy measure $\mu$ is lower-self- continuity.

Proof.

(necessity) if $\tilde{f}_n \mathop {\rightarrow }\limits ^{u.}\tilde{f}$, then $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.u.}\tilde{f}$, so $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (B_n )=0$, for arbitrary $A\,\in \,$F and $\{B_n \}\,\subseteq \,$F

Let

$$\begin{aligned} \tilde{f}_n (x)=\left\{ {\begin{array}{l} {\begin{array}{lll} 1&{} &{} {x\in B_n } \\ \end{array} } \\ {\begin{array}{lll} 0&{} &{} {x\notin B_n } \\ \end{array} } \\ \end{array}} \right. \end{aligned}$$

So

$$\begin{aligned} \mathop {\lim }\limits _{n\rightarrow \infty } \mu (\{x\left| {\left| {\text {Re}(\tilde{f}_n -0)} \right| } \right. \ge \varepsilon _1 , \left| {\text {Im}(\tilde{f}_n -0)} \right| \ge \varepsilon _2 \}\cap A)=\mathop {\lim }\limits _{n\rightarrow \infty } \mu (B_n )=0, \end{aligned}$$

where for arbitrary $\varepsilon _i >0, (i=1,2),\varepsilon =\varepsilon _1 +i\varepsilon _2$. So on $\text {A}$, if $\tilde{f}_n \mathop {\rightarrow }\limits ^{u.}0$, then $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.u.}0$.

Suppose for $\varepsilon _i <1$, then $\mu (A\backslash B_n )=\mu (\{x\left| {\left| {\text {Re}(\tilde{f}_n -0)} \right| } \right. <\varepsilon _1 , \left| {\text {Im}(\tilde{f}_n -0)} \right| <\varepsilon _2 \}\cap A)=\mu (A)$,

So $\mu$ is lower-self- continuity.

(Sufficiency): If $\tilde{f}_n \mathop {\rightarrow }\limits ^{u.}\tilde{f}$ on A, then $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (\{x\left| {\left| {Re(\tilde{f}_n -0)} \right| } \right. \ge \varepsilon _1 , \left| {Im(\tilde{f}_n -0)} \right| \ge \varepsilon _2 \}\cap A)=0$, where for arbitrary $\varepsilon _i >0,(i=1, 2),\varepsilon =\varepsilon _1 +i\varepsilon _2$. Let

$$\begin{aligned} B_n =\{x\left| {\left| {Re(\tilde{f}_n -\tilde{f})} \right| } \right. \ge \varepsilon _1 , \left| {Im(\tilde{f}_n -\tilde{f})} \right| \ge \varepsilon _2 \}\cap A, \end{aligned}$$

then $\{B_n \}\subseteq \text {A}$ and $\mathop {\lim }\limits _{n\rightarrow \infty } \mu (B_n )=0$. Due to $\mu$ is lower-self- continuity, so

$$\begin{aligned} \mu (A\cap \{x\left| {\left| {Re(\tilde{f}_n -\tilde{f})} \right| } \right. <\varepsilon _1 ,\left| {Im(\tilde{f}_n -\tilde{f})} \right| <\varepsilon _2 \})=\mu (A\backslash B_n )\rightarrow \mu (A). \end{aligned}$$

Therefore $\tilde{f}_n \mathop {\rightarrow }\limits ^{p.u.}\tilde{f}$ on A.

4 Conclusion

On the basis of the concept of complex Fuzzy measurable function in [2], we study its convergence theorem. It builds the certain foundation for the research of complex Fuzzy integral. Provide a strong guarantee for the complex fuzzy integral development, enrichment and development of complex fuzzy Discipline.