Suppose I have a function $x:\mathbb{R}\rightarrow\mathbb{R}$ such that is square-integrable:
$$\int_{-\infty}^\infty|x(t)|^2dt<\infty$$
Suppose also that $x(t)$ contains no higher frequencies than $B$ Hz. Then I can use the Nyquist theorem to recover $x(t)$ from its samples as long as these samples are spaced $T_s=\frac{1}{2B}$ seconds apart. I am trying to prove that the sequence $\{x(kT_s)\}_{k=-\infty}^\infty$ of samples of $x(t)$ is square-summable. Here is my approach:
$$\begin{eqnarray} \infty&>&\int_{-\infty}^\infty|x(t)|^2dt\\ &=&\int_{-\infty}^\infty\left(\sum_{k=-\infty}^\infty x(kT_s)\operatorname{sinc}(t/T_s-k)\right)\left(\sum_{l=-\infty}^\infty x(lT_s)\operatorname{sinc}(t/T_s-l)\right)dt\\ &=&\sum_{k=-\infty}^\infty \sum_{l=-\infty}^\infty x(kT_s)x(lT_s)\int_{-\infty}^\infty\operatorname{sinc}(t/T_s-k)\operatorname{sinc}(t/T_s-l)dt\\ &=&T_s^2\sum_{k=-\infty}^\infty |x(kT_s)|^2 \end{eqnarray}$$
However, I am unsure about the third step. Can I interchange the sums with integral there? Does Fubini's Theorem apply here?
Since the Integral of the sums converges, Then Fubini's theorem guarantees that the sum of the integral will also converge to the same answer.
A similar question was answered here.