# If $f:\mathbb{R}\to[0, \infty)$ (uniformly) continuous and $f \in L^1$, then $\lim_{x\to\pm\infty}f(x)=0$?

I'm learning about measure theory and need help with the following questions:

True or False (justify):

$(1)$ If $f:\mathbb{R}\to[0, \infty)$ measurable and $f \in L^1$, then $\lim_{x\to\pm\infty}f(x)=0$.

$(2)$ If $f:\mathbb{R}\to[0, \infty)$ continuous and $f \in L^1$, then $\lim_{x\to\pm\infty}f(x)=0$.

$(3)$ If $f:\mathbb{R}\to[0, \infty)$ uniformly continuous and $f \in L^1$, then $\lim_{x\to\pm\infty}f(x)=0$.

Since I'm having some difficulties for $(2)$ and $(3)$ I'm going to show my work for $(1)$.

My work for $(1)$:

The proposition is false. We consider the characteristic function of the rationals, that is the function $f:\mathbb{R}\to\mathbb{R}$ with

$$f(x)=\chi_{\mathbb{Q}}(x) = \begin{cases} 1 &\text{if x\in\mathbb{Q}} \\ 0& \text{if x\notin\mathbb{Q}} \end{cases}.$$

The characteristic function of the rationals is measurable. It is equal to zero almost everywhere. Therefore the function $f$ is integrable and and $\int f = 0$. However $\lim_{x\to\pm\infty}f(x)$ does not exist.

Edit: Considering the great answer of Umberto P., here's my complete answer for $(2)$:

The proposition is false. We start with the constant zero function. On the interval $[1, 3]$ we raise the graph as a triangle with a top at the point $(2, 1)$. The area under the first triangle, say $A_1$, is equal to $1$. From the point $(0, 3)$ we move one unit to the right on the real axis. On the interval $[4, 4.5]$ we raise again the graph as a triangle with a top at the point $(4.25, 1)$. The area under the second triangle $A_2$ is equal to $1/4$. We repeat this process indefinitely for triangles of height $1$. The resulting function is continuous, non-negative and therefore integrable. Moreover

$$A_1=1, A_2=\frac{1}{4}, A_3=\frac{1}{9}, A_4=\frac{1}{16}, \ldots \implies \int f = \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.$$

We see that the function integrates to $\frac{\pi^2}{6}$ but the graph has infinitely many bumps of height $1$. So the limit of $f$ does not exist.

## 1 Answer

For (2) consider a nonnegative continuous function whose graph has lots of bumps of height 1, but with the property that the total area under the bumps is finite.

On the other hand (3) is true. If the limit failed to exist or is nonzero, there is an $\epsilon > 0$ with the property that for any $x_0$ there is $x > x_0$ with $|f(x)| > \epsilon$. Since $f$ is uniformly continuous there is a $\delta > 0$ with the property that $|x-y| < \delta$ and $|f(x)| > \epsilon$ implies $|f(y)| > \epsilon/2$.

You can use this fact to find infinitely many disjoint intervals of length $2\delta$ on which $|f|$ exceeds $\epsilon/2$. This can be shown to contradict integrability.