If $f\in L^1(\Bbb R)$ is continuous then $f\in C_0(\Bbb R)$

Question

Suppose $f \in L^1(\Bbb R)$ is continuous. Then is it necessarily true that $f\in C_0(\Bbb R)$? It is easily seen to be true if $f$ is uniformly continuous, but I can't see whether it is true if $f$ is merely continuous. Thanks in advance.

Answer 1

Let $\displaystyle{(x)=1-n^{2}\mid x-n \mid} $ for $n-\dfrac {1}{n^{2}} \leq x \leq n+\dfrac{1}{n^{2}}$ for each $n$ and $0$ outside the intervals $\left(n-\dfrac{1}{n^{2}},n+\dfrac {1}{n^{2}}\right)$. Then $f$ is continuous and integrable but $f(n)=1$ for all $n$.

In fact $0 \leq f(x) \leq 1$ for all $x$ and $\displaystyle{\int \mid f(x)\mid dx \leq \sum \frac{2}{n^{2}} =\frac {\pi^{2}} {3}}$.

Answer 2

You can even have continuous unbounded real maps that are Lebesgue integrable. See here for an example.