Is the limit equal to zero?

Question

I know the statement " If $f\in$$L^1(\mathbb{R})$ is uniformly continuous, then $\lim_{|x|\to\infty}|f(x)|=0$ " is true. How about $f\in$$L^1(\mathbb{R})$ is continuous but not uniformly continuous ? Is true or false ?

Answer 1

Consider the function $$f=\sum_{n=1}^\infty \chi_{[n,n+\frac{1}{n^2}]}$$ whose integral is $$\int|f|\,d\mu=\int f\,d\mu=\sum_{n=1}^\infty\int\chi_{[n,n+\frac{1}{n^2}]}=\sum_{n=1}^\infty\mu([n,n+\tfrac{1}{n^2}])=\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$ so that $f\in L^1(\mathbb{R})$, but $f(n)=1$ for all integers $n\geq 1$, so that $\lim_{x\to\infty}f(x)$ isn't $0$ (in fact, it doesn't exist).

Now just make a continuous version of $f$ (with trapezoids, or bump functions, etc.)