Help understanding step in proof involving Lebesgue integrals over union of sets Let $A,B\subseteq\mathbb{R}^{2}$ be bounded and disjoint. Then:
$$
\int_{(A\cup B)^{n}}\mathbb{1}[\{x_{1},\dots,x_{n}\}\cap A\in F,\{x_{1},\dots,x_{n}\}\cap B\in G]\,dx_{1}\dots dx_{n}=\sum_{m=0}^{n}\frac{n!}{m!(n-m)!}\int_{A^{m}}\mathbb{1}[{x_{1},\dots,x_{m}}\in F]\int_{B^{n-m}}\mathbb{1}[{x_{m+1},\dots,x_{n}}\in G]\,dx_{1}\dots dx_{n}.
$$
My question: How do I see that the equality holds true?
Note that I've intentioanlly avoided addressing what exactly $F$ and $G$ are to spare some mathematical technicalities that I believe are irrelevant for this question.
 A: Recall $A\cap B =\emptyset$. So for $n=2$ and $f:(\mathbb{R}^2\times \mathbb{R}^2)\to \mathbb{R}$ integrable we have
$$\begin{aligned}\int_{(A\cup B)^2}f\,d(\lambda^2\times \lambda^2)&=\int_{A\cup B}\int_{A \cup B}f\,dx_1dx_2=\\
&=\int_{A\cup B}\bigg(\int_Af\,dx_1+\int_Bf\,dx_1\bigg)dx_2=\\
&=\int_{A}\bigg(\int_Af\,dx_1+\int_Bf\,dx_1\bigg)dx_2+\int_{B}\bigg(\int_Af\,dx_1+\int_Bf\,dx_1\bigg)dx_2=\\
&=\int_A\int_Af\,dx_1dx_2+\int_A\int_Bf\,dx_1dx_2+\int_B\int_Af\,dx_1dx_2+\int_B\int_Bf\,dx_1dx_2\end{aligned}$$
Now in this case
$$\begin{aligned}f(x_1,x_2)=\mathbf{1}_{\{(x_1,x_2):(x_1,x_2)\cap A \in F\}}(x_1,x_2)\cdot \mathbf{1}_{\{(x_1,x_2):(x_1,x_2)\cap B \in G\}}(x_1,x_2)\end{aligned}$$
Notice we integrate always on either $A$ or $B$, which are disjoint. So for example
$$\begin{aligned}x_1 \in A,\,x_2 \in B&\implies (x_1,x_2)\cap A=x_1,\,(x_1,x_2)\cap B=x_2\\
&\implies f(x_1,x_2)=\mathbf{1}_{\{(x_1,x_2):x_1\in F,x_2\in G\}}(x_1,x_2)\\
&\implies f(x_1,x_2)=\mathbf{1}_F(x_1)\cdot\mathbf{1}_G(x_2)\end{aligned}$$
Therefore
$$\begin{aligned}\int_A\int_Af\,dx_1dx_2&=\int_A\int_A\mathbf{1}_F(x_1)\cdot \mathbf{1}_F(x_2)\,dx_1dx_2=\int_{A^2}\mathbf{1}_{\{x:x\subseteq F\}}(x)d(\lambda^2\times \lambda^2)\\
\int_A\int_Bf\,dx_1dx_2&=\int_A\int_B\mathbf{1}_G(x_1)\cdot\mathbf{1}_F(x_2)dx_1dx_2=\int_A\mathbf{1}_F(x_2)\int_B\mathbf{1}_G(x_1)dx_1dx_2\\
\int_B\int_Af\,dx_1dx_2&=\int_B\int_A\mathbf{1}_F(x_1)\cdot\mathbf{1}_G(x_2)dx_1dx_2=\int_B\mathbf{1}_G(x_2)\int_A\mathbf{1}_F(x_1)dx_1dx_2\\
\int_B\int_Bf\,dx_1dx_2&=\int_B\int_B\mathbf{1}_G(x_1)\cdot\mathbf{1}_G(x_2)dx_1dx_2=\int_{B^2}\mathbf{1}_{\{x:x\subseteq G\}}(x)d(\lambda^2\times \lambda^2)\end{aligned}$$
Note that in this case
$$\int_A\int_Bf\,dx_1dx_2=\int_B\int_Af\,dx_1dx_2$$
So finally with great abuse of notation
$$\begin{aligned}\int_{(A\cup B)^2}f\,d(\lambda^2\times \lambda^2)&=\int_{A^2}\mathbf{1}_{\{x:x\subseteq F\}}(x)d(\lambda^2\times \lambda^2)+2\int_B\mathbf{1}_G(x_2)\int_A\mathbf{1}_F(x_1)dx_1dx_2+\int_{B^2}\mathbf{1}_{\{x:x\subseteq G\}}(x)d(\lambda^2\times \lambda^2)\\
&=\sum_{m=0}^2\binom{2}{m}\int_{A^m}\mathbf{1}_{\{(x_1,...,x_m):(x_1,...,x_m)\in F\}}(x_1,...,x_m)\int_{B^{2-m}}\mathbf{1}_{\{(x_{m+1},...,x_2):(x_{m+1},...,x_2)\in G\}}(x_{m+1},...,x_2)dx_1dx_2\end{aligned}$$
I believe that the general case with $n>2$ can be derived in a similar manner.
