Lebesgue integrable function without compact support.

Question

Suppose $f \in L^1(\mathbb{R}, m)$, where $m$ is Lebesgue measure. By definition we have

$$\int_{\mathbb{R}} f dm < \infty$$

Does $f$ have compact support? This makes sense, but I don't know if it's true.

Answer 1

This is false. Since the Lesbegue integral ignores countable sets, and $\mathbb{R}$ is seperable, you can take any Lesbegue integrable function, define it to be nonzero on some countable dense set and you have a set that's not compactly supported.