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I have been working in the following problem but I'm having no luck with it.

Let $f\colon \mathbb R\to \mathbb R^+$ be a measurable function. It must be shown that $$(*)\qquad \int\limits_{[0, 1]} \sum_{k\in \mathbb Z} f(x+k)\,dx=\int\limits_{\mathbb R}f(x)\,dx.$$ Moreover, what can be said of the series $\sum\limits_{k\in \mathbb Z} f(x+k)$ provided that $f $ is integrable?.

I've been trying to express the LHS of $(*)$ as an expression with sums of indicator functions, but it only gets bigger and I don't see how it transforms into the RHS. I like the problem, it is visually pleasing. It must not be that hard, but I just don't see which one is the trick that works.

Can you help me? Thanks in advance!

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  • $\begingroup$ Looks like you would like to exchange limit and summation... $\endgroup$ – Niklas Sep 10 '17 at 22:35
  • $\begingroup$ @ Niklas Hebestreit ... Following your remark, the OP is invited to see if (*) with an exchange between $\sum$ and $\int$ (provided it is "legal") in the LHS gives a result that is easier to establish (think to a change of variable). $\endgroup$ – Jean Marie Sep 10 '17 at 22:40
  • $\begingroup$ @JeanMarie, NiklasHebestreit I think I got it! $\endgroup$ – EternalBlood Sep 10 '17 at 23:14
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If $\displaystyle g(x)= \sum_{k\in \mathbb Z} f(x+k)\,dx,$ then $g$ is measurable because each $f_k(x)=f(x+k)$ is.

Moreover, if we take $y=x+k$,

$$\int\limits_{[0, 1]} g(x)\,dx=\int_{\mathbb R} g(x)\chi_{[0, 1]}(x)\,dx =\sum_{k\in \mathbb Z}\int_{\mathbb R} f(x+k)\chi_{[0, 1]}(x)\,dx =\sum_{k\in \mathbb Z}\int_{\mathbb R} f(y)\chi_{[0, 1]}(y-k)\,dy=\sum_{k\in \mathbb Z}\int_{\mathbb R} f(y)\chi_{[k,k+ 1]}(y)\,dy=\sum_{k\in \mathbb Z}\int_{[k, k+1]} f(y) \,dy=\int_{\mathbb R} f(y)\, dy.$$

I exchanged several things where needed, but they are all valid

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