Background
Let $G$ be a locally compact abelian group and $M(G)$ the Banach space of all complex Radon measures on $G$. The convolution of two complex Radon measures $\mu, \lambda\in M(G)$ is defined by specifying a linear functional on $C_0(G)$ (and invoking Riesz representation theorem) as follows:
$$\int_G\phi\,\mathrm{d}(\mu*\nu)=\int_G\int_G\phi(xy)\,\mathrm{d}\mu(x)\,\mathrm{d}\nu(y)$$ for $\phi\in C_0(G)$.
However in Rudin's Fourier Analysis on Groups (see page 14 for details) to prove that the convolution of measures are associative (see Theorem 1.3.2(b)) he uses the following definition convolution of two measures of $M(G)$ as follows,
Definition 1. Let $\mu,\lambda\in M(G)$ then the set function $\mu\ast \lambda$ is defined as follows, $$(\mu \ast \lambda)(E):=(\mu\times \lambda)(E_{(2)})$$for each Borel set $E$ in $G$ where $\mu\times \lambda$ denotes the product measure on $G\times G$ and, $$E_{(2)}:=\{(x,y)\in G\times G:x+y\in E\}$$
To prove associativity of convulotion, he generalizes the above definition to $n$-measures as follows,
Definition 2. Let $\mu_1,\ldots,\mu_n\in M(G)$ then the set function $(\mu_1\ast\ldots\ast\mu_n)$ is defined as follows, $$(\mu_1\ast\ldots\ast\mu_n)(E):=(\mu_1\times\ldots\times\mu_n)(E_{(n)})$$for each Borel set $E$ in $G$ where $(\mu_1\times\ldots\times\mu_n)$ denotes the product measure on $G^n$ and, $$E_{(n)}:=\{(x_1,\ldots,x_n)\in G^n:x_1+\ldots+x_n\in E\}$$
and simply says that the associativity of measures follows by Fubini's Theorem.
Questions
Let's take $n=3$ and consider the then the first problem that I am facing is regarding the definition of $\mu_1\ast (\mu_2\ast \mu_3)$ in accordance with Definition 2.
Sure,by applying Definition 1 we can say that,$$(\mu_1\ast (\mu_2\ast \mu_3))(E):=(\mu_1\times (\mu_2\ast \mu_3))(E_{(2)})$$but then what? Similar is the problem with the definition of $(\mu_1\ast \mu_2)\ast \mu_3$.
How can we use Fubini's theorem to prove the result?
Let $\lambda,\mu, \nu\in M(G)$. Let $\alpha:G\times G\rightarrow G$ be the group operation +. Let $E$ be a Borel measurable set. \begin{align} (\lambda*\mu)*\nu(E)&=(\lambda*\mu)\times \nu(\alpha^{-1}(E))\\ &=\iint \chi_E(x+z)d(\lambda*\mu)(x)d\nu(z) \end{align} Now for each $z\in G$, let $E-z=\{e-z:e\in E\}$. We then have \begin{align} \int \chi_E(x+z)d(\lambda*\mu)(x)&=\int \chi_{E-z}(x)d(\lambda*\mu)\\ &=(\lambda*\mu)(E-z)\\ &=(\lambda\times \mu)(\alpha^{-1}(E-z))\\ &=\iint\chi_{E-z}(x+y)d\lambda(x)d\mu(y)\\ &=\iint\chi_{E}(x+y+z)d\lambda(x)d\mu(y)\\ \end{align} We conclude that $$(\lambda*\mu)*\nu(E)=\iiint\chi_E(x+y+z)d\lambda(x)d\mu(y)d\nu(z)$$ Similarly one can prove the following equality by using Fubini's Theorem: $$\lambda*(\mu*\nu)(E)=\iiint\chi_E(x+y+z)d\lambda(x)d\mu(y)d\nu(z)$$ which suggests that convolution of measures is associative.