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Prove that if f(x) is continuous and absolutely integrable on $[a,\infty)$ then $\lim\limits_{x\to\infty}f(x)=0$.

I tried proving it in the following way: First we need to prove the existence of the border. Let $\epsilon>0$. since f is continuous exists $\delta>0$ s.t $\forall x_0\in[a,\infty)$ if $|x-x_0|<\delta,|f(x)-f(x_0)|<\epsilon$. We need to find $S>0$ s.t $\forall x>S ||f(x)|<\epsilon$. Can we take $S=\delta+x_0$ (I hesitate since it mustn't shouldn't depand on $x_0$). After that if without loss of generality $\lim\limits_{x\to\infty}f(x)=L>0$, it follows $|f(x)|>L>0$ and by integrating both sides it follows the integral diverges in contradiction to the details and we finished.

Am I right? Which S should I take?

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  • $\begingroup$ This isn't true. Do you have the hypothesis that $f$ is uniformly continuous on $[a,\infty]$? This post has proofs for this case. $\endgroup$ – David Mitra Aug 17 '13 at 9:29
  • $\begingroup$ Why that's not correct? $\endgroup$ – user65985 Aug 17 '13 at 9:42
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Whether or not this is true depends on the meaning of your notation. If by $[a,\infty]$ you mean an interval that includes $\infty$ (ie. that $f$ has a limit at $\infty$), then it is not hard to see that if the limit at $\infty$ is anything but $0$, then $\int_a^\infty f(x)\,dx$ must diverge. That is, say that $\lim_{x \to \infty} |f(x)|=L$. Then take $S$ such that $||f(x)|-L| < \frac L2$ for all $x > S$. In particular, this means that $|f(x)| < L-\frac L2 = \frac L2$ for all $x > S$. Thus, $$ \int_S^T |f(x)|\,dx \geq (T-S)\frac L2 $$ for $T > S$, and thus, the integral diverges.

On the other hand, if you aren't assuming that $f$ has a limit at infinity (in which case, I would say that $f$ is continuous on $[a,\infty)$), then here is a counterexample:

Let $f$ be any nonnegative-valued continuous function on $[0,\infty)$ such that $f(n)=1$ for each natural number $n$ and $f(x) = 0$ outside of

$$ \bigcup_{n \in \mathbb N} (n-\frac1{2^n}, n+\frac1{2^n}). $$

(For example, $f$ could be piecewise linear.) Then $\int_0^N f(x)\,dx \leq \sum_{n=1}^n \frac2{2^n} < 1$, and from this, and the fact that $f$ is nonnegative-valued, we see that $f$ is (absolutely) integrable. But, since $f(n)=1$ for each natural number $n$, and between each pair of natural numbers, there is a point where $f$ takes on the value $0$, $\lim_{x \to \infty} f(x)$ does not exist.

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  • $\begingroup$ I meant $[a,\infty)$. $\endgroup$ – user65985 Aug 17 '13 at 15:38

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