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?
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. $[\mu]$.
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 $L^2(\mathbb{R})$.
