How to rewrite integral of L^2 function and conclude about convergence of series

Question

I've read the following assertions:

" Suppose $f \in L^2(\mathbb{R}). $ Then $$ \int_{-\frac{1}{2}}^{\frac{1}{2}} \sum_{k \in \mathbb{Z}} \vert f(x+k) \vert^2 dx = \int_{-\infty}^{\infty} \vert f(x) \vert^2 dx < \infty. $$ Thus, $ \sum_{k \in \mathbb{Z}} \vert f(x+k) \vert^2 < \infty $ for $a.e. x\in \mathbb{R}.$ "

Why exactly does the given equality between the two integrals hold and what property or theorem gives the convergence of the sum?

Answer 1

You can always interchange the sum and the integral becasue of non-negativity. Now $\int_{-\frac 1 2} ^{\frac 1 2 } |f(x+k)|^{2}dx=\int_{k-\frac 1 2}^{k+\frac 1 2} |f(y)|dy$ by the substitution $y=x+k$. Now sum over all $k$. Note that intervals $[k-\frac 1 2, k+\frac 1 2)$ from a partition of the real line.

If $f$ is a non-negative measurable function and $\int f d\mu <\infty$ then $f <\infty$ a.e. $[\mu]$.

Answer 2

We can write any real number $x \in \mathbb{R}$ as a sum $k + \epsilon$, with $k \in \mathbb{Z}, \epsilon \in \left[-\frac12, \frac12\right)$ in a unique way.

That's what the sum on the left does explicitly. If you replace the variable $x$ with $\epsilon$ in the left-hand-side, it should be even more clear.

The integral being finite is given as part of the definition of a function in $L^2(\mathbb{R})$.