Bochner's theorem

Question

I'm reading Bochner's theorem. Now I'm having problem with part on the third page: $\int_{s\in [0,T], s+u\in [0, T]} ds=1-\frac{|u|}{T}$? How to deduce it?

Any help is welcome. Thanks in advance.

Answer 1

What is true is that for $|u|\leq T$, $$ \int_{s\in[0,T],s+u\in[0,T]} \,ds = T - |u|, $$ and in those lecture notes notice that a factor of $T$ is dropped, which gives you $(1-|u|/T)$. Moreover we do have $|u|\leq T$ because $u$ is being integrated from $-T$ to $T$.

To do this integral, just observe that when $u>0$, $$ \{s\in [0,T] \mid s+u\in [0,T]\} = [0,T-u] $$ and likewise when $u<0$, $$ \{s\in [0,T] \mid s+u\in [0,T]\} = [u,T]. $$