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Let $\{(X_i,\Sigma_i, \mu_i)\}_{i=1}^\infty$ be a collection of $\sigma$-finite measure space.
For each $i$, let $f_i$ be a $\mu_i$-measurable nonnegative function.
Define the product measure space $$ \{X,\Sigma,\mu\}=\{\prod_{i=1}^\infty X_i, \prod_{i=1}^\infty \Sigma_i, \prod_{i=1}^\infty \mu_i \} $$ and the function $g: X \to \mathbb R_+$ $$ g(x)=\prod_{i=1}^\infty f_i(x_i). $$

Is Tonelli theorem applicable to $$ \int_X g(x) d\mu(x) \; ? $$

That is, order of integration can be interchanged at will to get $$ \int_X g(x) d\mu(x) = \prod_{i=1}^\infty \int_{X_i} f(x_i)d\mu_i(x). $$

I would say yes, by inductively applying Tonelli theorem pairwise, then taking the limit.
But I'm not sure it's correct.

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It seems to me that this is more a question about changing $\prod$ and $\int$, as: Let $g_i\colon X \to [0,\infty]$ given by $g_i := f_i \circ \pi_i$ ($\pi_i$ the projection), then $g = \prod_i g_i$ and $\int_X g\, d\mu = \int_{X_i} f_i\, d\mu_i$ by definition of $\mu$. So if the convergence of $\prod_{i=1}^N g_i \to g$ is monotone or bounded by an integrable function, your result should hold ... – martini Oct 26 '12 at 11:19
It seems that in general the function $g$ is not measurable with respect to the product sigma-algebra $\otimes_i\Sigma_i$, a fact which would make pointless the rest of the question. – Did Oct 26 '12 at 12:17

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