# Application of the monotone convergence theorem to a product measure

I am trying to understand the poof of a theorem in a book. Consider the product measure $\mu = \times_{n = 1}^{\infty} \mu_n$ on $\mathbb{R}^{\infty}$ where $\mu_n$ are finite measures on $\mathbb{R}$. Let $x \in \ell_2$. How does one conclude by the monotone convergence theorem that $\int_{\mathbb{R}^{\infty}} \sum_{k = 1}^{\infty} x_k^2 \mu \left( \mathrm{d} x \right) = \sum_{k = 1}^{\infty} \int_{\mathbb{R}} x_k^2 \mu_k \left( \mathrm{d} x_k \right)$

My attempt:

The sequence $s_n = \sum_{k = 1}^n x_k^2$ is monotone since $s_n \leqslant s_{n + 1}$ for $n \in \mathbb{N}$. \begin{eqnarray*} \int_{\mathbb{R}^{\infty}} \sum_{k = 1}^{\infty} x_k^2 \mu \left( \mathrm{d} x \right) & = & \int_{\mathbb{R}^{\infty}} \lim_{n \rightarrow \infty} \sum_{k = 1}^n x_k^2 \mu \left( \mathrm{d} x \right)\\ & & \text{by the MCT}\\ & = & \lim_{n \rightarrow \infty} \int_{\mathbb{R}^{\infty}} \sum_{k = 1}^n x_k^2 \mu \left( \mathrm{d} x \right)\\ & = & \lim_{n \rightarrow \infty} \int_{\mathbb{R}^n} \sum_{k = 1}^n x_k^2 \mu_1 \times \cdots \times \mu_n \left( \mathrm{d} x_1 \cdots \mathrm{d} x_k \right)\\ & = & \lim_{n \rightarrow \infty} \sum_{k = 1}^n \int_{\mathbb{R}^n} x_k^2 \mu_1 \times \cdots \times \mu_n \left( \mathrm{d} x_1 \cdots \mathrm{d} x_k \right)\\ & = & \lim_{n \rightarrow \infty} \sum_{k = 1}^n \int_{\mathbb{R}} x_k^2 \mu_k \left( \mathrm{d} x_k \right)\\ & = & \sum_{k = 1}^{\infty} \int_{\mathbb{R}} x_k^2 \mu_k \left( \mathrm{d} x_k \right) \end{eqnarray*} Is this the correct way of writing the proof in a very detailed fashion? Am I jumping some logical hurdles here (for instance when writing $\mu(\mathrm{d}x)$ at the beginning?

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The second line doesn't look quite right. Not that the sequence $f_n \colon\mathbb R^{\infty} \to [0,\infty)$ given by $f_n(x) = \sum_{k=1}^n x_k^2$ is monotone. So you have $$\int_{\mathbb R^\infty} \lim f_n(x) \,d\mu = \lim \int_{\mathbb R^\infty} f_n(x)\, d\mu$$ with an integral over $\mathbb R^\infty$. If the $\mu_k$ are probability measures (which I will assume in the following and you also seem to do so), we have by Fubini that is equals $$\lim \sum_{k=1}^n \int_{\mathbb R} x_k^2 d\mu_k = \sum_{k=1}^\infty \int_{\mathbb R} x_k^2 \, d\mu_k.$$
@learner You just have Fubini for $n+1$ factors $\mathbb R^\infty \cong \mathbb R \times \cdots \mathbb R \times \mathbb R^\infty_{\text{rest}}$ where $\mathbb R^\infty_{\text{rest}}$ has the coordinates $(x_{n+1},x_{n+2},\ldots)$, as the integrand does not depend on these, we have the integral over this equal to the constant integrand, as $\mu_{\text{rest}} = \bigotimes_{k=n+1}^\infty \mu_k$ is a probability measure. – martini Nov 22 '12 at 7:29