Let $f\in L^1(\mathbb{R}^n)$ und $F(x)= \sum_{k \in \mathbb{Z}^n} f(x-k)$ be it's periodization.
Does $F$ converge absolutely or uniformly?
I am asking because I lately saw something like this:
$$\int_{(0,1)^n}\sum_{k \in \mathbb{Z}^n} f(x-k) = \sum_{k \in \mathbb{Z}^n} \int_{(0,1)^n} f(x-k) $$ and I was wondering why this is correct.
$F$ converges neither absolutely nor uniformly. In fact, it might not converge at all!
Let $$f(x)=\sum_{j\in\mathbb{Z}}{1_{|x-j|<2^{-j}}}$$ Then $\|f\|_1=\sum_{j\in\mathbb{Z}}{2\cdot2^{-j}}=6$, but $$F(0)=\sum_{k\in\mathbb{Z}}{\sum_{j\in\mathbb{Z}}{1_{|0+k-j|<2^{-j}}}}\geq\sum_{j=k\in\mathbb{Z}}{1}=\infty$$
Of course, if $F$ is only considered as an element of $L^1(\mathbb{R})$, then this is no problem. But since the $L^1$-norm is translation invariant, it is easy to see that each term of the sum defining $F$ changes the $F$-norm by $\|f\|_1=6$, so that $F$ fails to converge in $L^1(\mathbb{R})$ too.
The interchange of limits you observe is best described as a convenient rewriting of a single integral, split into countably-many disjoint parts: $$\sum_{k\in\mathbb{Z}^n}{\int_{[0,1)^n}{f(x-k)\,dx}}=\sum_{k\in\mathbb{Z}^n}{\int_{-k+[0,1)^n}{f(x)\,dx}}=\int_{\mathbb{R}^n}{f(x)\,dx}$$ For this reason, $F$ converges in $L^1(K)$ for any compact $K$.