For each $n\in\Bbb Z$, let $f\_n\colon [n,n+1]\to\Bbb R$ be a function such that $f\_{n}(n+1)=f\_{n+1}(n+1)$ and $f\_n$ is $0$ on $\Bbb R\setminus [n,n+1]$. Now, define $f$ as follows $$f=\sum\_{n\in\Bbb Z} f\_n$$ Clearly, $f$ is a function form $\Bbb R$ to $\Bbb R.$ My question is

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> Can I think for $f$ as $f=\bigcup\_{n\in\Bbb Z} f\_n.$?
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I think they are the same. Is that right? Any idea.