Proving Tonelli's Theorem for $n$ Factors Am trying to prove the following extension of Tonelli's theorem:
Proposition. Let $(\Omega_j,\mathcal{A}_j,\mu_j)$ $j=1,\dots,n$ be $\sigma$-finite measure spaces. Let $f\to[0,\infty]$ be an $\mathcal{A}_1\otimes \dots\otimes\mathcal{A}_n$ measurable function on $\Omega_1\times\dots\times\Omega_n$. Then for every permutation $j_1,\dots,j_n$ of $1,\dots,n$ we have
$$\int f(\omega_1,\dots,\omega_n) \,d (\mu_1 \otimes \dots \otimes \mu_n)=\int \dots \int f(\omega_1,\dots,\omega_n)\,d\mu_{j_1}\dots d\mu_{j_n}$$
where each integral on th RHS is measurable with respect to the product of the $\mathcal{A}_j$ corresponding to coordinates in which integration has not yet occured. My book says it's a simple induction but somehow my proof seems complicated.
I believe it is sufficient to consider the case of the identity permutation. This is because we have the equality
$$\int f(\omega_1,\dots,\omega_n) \,d (\mu_1 \otimes \dots \otimes \mu_n)=\int f(\omega_{1},\dots,\omega_{n}) \,d (\mu_{j_1} \otimes \dots \otimes \mu_{j_n})$$
see here. In other words, it does not matter whether we regard $f$ as a function on $\Omega_1\times\dots\times\Omega_n$ or on $\Omega_{j_1}\times\dots\times\Omega_{j_n}$.
Is this correct? The comments seem to indicate that are multiple possible approaches here. Any proof outline is greatly appreciated.
 A: Let's write down a statement of Tonelli's theorem, just to make everything clear.
Let $(X, \mathcal M, \mu)$ and $(Y, \mathcal N, \nu)$ be $\sigma$-finite measure spaces and $f: X \times Y \to [0,\infty]$ be $\mathcal M \otimes \mathcal N$ measurable. Then:
$$\int f d(\mu \times \nu) = \int \left(\int f(x_1, x_2) d\nu(x_2)\right) d\mu(x_1) = \int \left(\int f(x_1, x_2) d\mu(x_1)\right) d\nu(x_2).$$
Alright now we can write down a proof of your statement. Let $(X_j, \mathcal M_j, \mu_j)$ be a finite collection of $\sigma$-finite measure spaces and let $f : \prod_{j=1}^n X_j \to [0,\infty]$ be $\bigotimes_{j=1}^n \mathcal M_j$ measureable.
Note that $\bigotimes_{j=1}^n \mathcal{M}_j = \mathcal M_1 \otimes \bigotimes_{j=2}^n \mathcal M_j$, that $\mu_1 \times \mu_2 \times \cdots \times \mu_n = \mu_1 \times (\mu_2 \times \cdots \times \mu_n)$, and that $(\prod_{j=2}^n X_j,  \bigotimes_{j=2}^n \mathcal M_j, \mu_2 \times \cdots \mu_n)$ is $\sigma$-finite (this is direct from proving the product of $\sigma$-finite measure spaces is $\sigma$-finite and induction).
Thus repeated application of the first application of Tonelli's theorem above (i.e. an induction) gives that:
$$\int f d(\mu_1 \times \cdots \times \mu_n) = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_1(x_1)\right) \cdots \right) d\mu_n(x_n).$$
Now we show inductively show that, under the assumptions of our proposition, that:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_1(x_1)\right) \cdots \right) d\mu_n(x_n) = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_{\sigma(1)}(x_{\sigma(1)})\right) \cdots \right) d\mu_{\sigma(n)}(x_{\sigma(n)})$$
for any permutation $\sigma \in S_n$.
It is clearly true for $n=1$.
Suppose it has been shown for $n$, then choose $\sigma \in S_{n+1}$. Then define $\tau \in S_n$ inductively by $\tau(1) = \sigma(1)$ if $\sigma(1) \ne n+1$ else $\sigma(2)$ and $\tau(j+1) = \sigma(\sigma^{-1}(\tau(j))+1)$ if $\sigma(\sigma^{-1}(\tau(j))+1) \ne n+1$ else $= \sigma(\sigma^{-1}(\tau(j))+2)$.
The result is that $\tau$ arranges $1,...,n$ in the same order as $\sigma$. Then applying the inductive hypothesis with $\tau$ to the interior integral for each $x_{n+1}$:
$$\int \left (\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_1(x_1)\right) \cdots \right) d\mu_n(x_n) \right) d\mu_{n+1}(x_{n+1}) = \int \left(\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_{\tau(1)}(x_{\tau(1)})\right) \cdots \right) d\mu_{\tau(n)}(x_{\tau(n)})\right) d\mu_{n+1}(x_{n+1}).$$
Then since $\tau$ put $1,...,n$ into the same order as $\sigma$, all that is left to get $1,...,n+1$ into the order induced by $\sigma$ is to insert $d\mu_{n+1}(x_{n+1})$ in the right spot, for which it suffices to show that two adjacent $d\mu_i(x_i)$ and $d\mu_j(x_j)$ can be commuted (then repeatedly commuting $d\mu_{n+1}(x_{n+1})$ left until it is in the right spot finishes the proof).
This we will now do. Claim:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_i  \bigg) d\mu_j \cdots \right) d\mu_b(x_n) = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_j  \bigg) d\mu_i \cdots \right) d\mu_b(x_n).$$
But this is just a direct application of Tonelli's theorem, since it suffices to show that:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_i  \right) d\mu_j  = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_j  \right) d\mu_i,$$
and we have:
$$\int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_i  \right) d\mu_j  = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \right) d\mu_j \times \mu_i = \int \left(\cdots \left(\int f(x_1,...,x_n) d\mu_a(x_1)\right) \cdots \bigg) d\mu_j  \right) d\mu_i.$$
Putting it together completes the proof.
Note: Alternatively, instead of all that $\tau$ stuff, we can use the final claim to show that the set of permutations of the measures is a subgroup containing consecutive permutations: $(i, i+1)$ and then prove that $(i, i+1)$ generates $S_n$, which effectively what I did in the "$\tau$-section", though it might be a bit confusing.
