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?