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$f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$

Given that $f$ is uniformly continuous function on real and is integrable with respect to lebesgue measure, we need to show $f(x)\rightarrow 0$ as $x\rightarrow\infty$.

I have no clue for this problem, thank you for help.


marked as duplicate by rschwieb, Davide Giraudo, Did, Hagen von Eitzen, Stefan Hansen Jan 27 '13 at 13:55

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    $\begingroup$ Basically, the idea is this. Let $\epsilon>0$. Then there exists a $\delta>0$ such that the entire graph of $f$ can be covered by squares $2\delta$ long and $2\epsilon$ high, touching edge to edge. If $f$ did not converge to zero, then there would have to exist infinitely many "bumps" that reach a certain height, and if you choose $\epsilon$ small enough, there will be space left between the box and $x=0$. This little space occurring infinitely many times would contradict the integrability of $f$. $\endgroup$ – rschwieb Jan 27 '13 at 12:59
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    $\begingroup$ Prove that if $f$ is uniformly continuous and $f(x)\nrightarrow 0$ as $x\to \infty$, then $f$ is not lebesgue integrable. $\endgroup$ – aliakbar Jan 27 '13 at 13:05
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    $\begingroup$ I recommend that you try to construct a function which is continuous, but not uniformly continuous, integrable and $f(x) \nrightarrow 0$ for $x \to \infty$. The answers contain some hints for this. $\endgroup$ – Martin Jan 27 '13 at 13:14
  • $\begingroup$ $f(x)=x^2$ works....? $\endgroup$ – Marso Jan 27 '13 at 13:17
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    $\begingroup$ That's not integrable. $\endgroup$ – Martin Jan 27 '13 at 13:17

Suppose that $f(x) \not \to 0$ as $x \to \infty$. This means that: $$ \exists \epsilon > 0 : \forall M > 0, \exists x > M : x > M \implies \left|f(x)\right| > \epsilon $$

Since $f$ is uniformly continuous, for this $\epsilon$ there is $\delta > 0$ satisfying: $$ \forall x, y > 0 : |x - y| < \delta \implies \left|f(x) - f(y)\right| < \epsilon/2 $$

Putting the two together, we find that: $$ \forall M > 0 : \exists x > M : y \in (x -\delta, x +\delta) \implies f(y) > \epsilon /2 $$

Thus: $$ \int_{x-\delta}^{x+\delta} \left|f(y)\right| \,dy \ge \dfrac{\delta \epsilon}{2} $$

This process can be repeated as many times as we like. We start by finding $x_0 > 1$. Next, we find $x_1 > x_0 + 2\delta$. And we continue by finding $x_n > x_{n-1} + 2\delta$. None of the intervals $(x_n - \delta, x_n + \delta)$ overlap. Thus: $$ \int_0^\infty \left|f(y)\right| \,dy \ge \dfrac{\delta \epsilon n}{2} \quad \forall n \in \mathbb{N}, n > 0 $$

We conclude that $f$ is not Lebesgue integrable.


Expanding my comment into an answer, since the ones at the duplicate I found are comparatively hard to read.

Since $f$ can be rewritten as the difference of two nonnegative uniformly continuous functions, we can proceed by showing this holds for such functions $f\geq 0$.

The definition of uniform continuity says that for any $\epsilon>0$, there exists $\delta>0$ such that the entire graph can be covered with rectangles of width $2\delta$ and height $2\epsilon$, placed end to end.

Suppose $f(x)$ did not converge to zero on the right. Then there exists a $k>0$ such that $f(x)$ jumps above $k$ infinitely many times. Set $\epsilon=k/4$ and find the associated $\delta$. By this choice of $\epsilon$, there will always be a box of width $2\delta$ and height at least $k/2$ fully under the curve at each point where $f(x)$ goes above $k$.

Such boxes have area at least $\delta k$, and since there are infintely many of them under the curve, this would contradict Lebesgue integrability. Hence, $f$ tends to zero on the right.


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