Why Does a Real-Valued Function on $[0,\infty)$ Have Limit $0$ at Infinity?

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Here is this week's POTW:

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Explain why a uniformly continuous, Lebesgue integrable, real-valued function on $[0,\infty)$ has limit $0$ at infinity.

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No one answered this week's problem. You can read my solution below.

Spoiler

Let $f$ be such a function. Let $\epsilon > 0$. Since $f$ is uniformly continuous, there exists $\delta > 0$ such that for all $x,y$, $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$. Since $f$ is Lebesgue integrable on $[0,\infty)$, there exists $M > 0$ such that for all $x,y$, $x > y > M$ implies $\int_y^x |f(t)|\, dt < \epsilon$. Given $x > M$, there exists $c \in (x,x+\delta)$ such that $|f(c)|\delta = \int_x^{x+\delta} |f(t)|\, dt$, and $|f(c)|\delta < \epsilon$. Since $|x - c| < \delta$, then $|f(x) - f(c)| < \epsilon$. Hence

$$|f(x)| \le |f(x) - f(c)| + |f(c)| < \epsilon + \frac{\epsilon}{\delta} = \left(1 + \frac{1}{\delta}\right)\epsilon.$$

Since $\epsilon$ was arbitary, $\lim\limits_{x\to \infty} f(x) = 0$.