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I am looking for a function $f:[a, +\infty)\to\mathbb{R}$ with $a\in\mathbb{R}$ such that $f(x)$ satisfies all the following conditions:

  1. $f(x)$ is Riemann Integrable on $[a,m]$ for all $m>a$;
  2. $\lim_{x\to +\infty}f(x)$ does not exist;
  3. $f(x)$ is an unbounded function on $[b, +\infty)$ for all $b>a$;
  4. $\int_{a}^{+\infty}f(x)dx\in\mathbb{R}$.

Here's my attempt:

Let

$f(x)=\begin{cases}n&\mbox{ if }x\in \mathbb{N}\\ 0&\mbox{ otherwise}\end{cases}$

and $[a, +\infty)=[1, +\infty)$

  1. $f(x)$ is Riemann integrable over $[1, m]\mbox{ for all }m>1$, because is a continuous function with a finite number of discontinuities;

  2. $\lim_{x\to +\infty}f(x)$ does not exists, in fact I can consider two sequences $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ such that

    $\forall n\in\mathbb{N}, \ x_n\in\mathbb{N}\mbox{ and }\lim_{n\to +\infty}x_n=+\infty$

    $\forall n\in\mathbb{N}, \ y_n\notin\mathbb{N}\mbox{ and }\lim_{n\to +\infty}y_n=+\infty$

and I can prove that $\lim_{n\to +\infty}f(x_n)=+\infty\mbox{ but }\lim_{n\to +\infty}f(y_n)=0$.

  1. $f(x)$ is unbounded over $[b,+\infty)\mbox{ for all }b>1$, in fact $\lim_{n\to +\infty}f(x_n)=+\infty$.
  2. Now I think, but it's just a congecture, that $\int_{1}^{+\infty}f(x)dx=0$, is this true? How can I explain this?

Thank you.

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    More interesting: the same, with $f$ smooth. – 2017-01-24
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    I second the "now make $f$ smooth" requirement - there are some easily-defined functions which it is very good for you to know about as examples and counterexamples in analysis, and this will help you find at least one of them. (You might find it easier first to make $f$ merely *continuous* rather than smooth.) – 2017-01-24
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    I think that if $f(x)$ is smooth function (i.e derivable function?) over $[a, +\infty)$ than $\int_{a}^{+\infty}f(x)dx$ surely does not converge. Right? Sorry, English is not my first language, maybe I misanderstood your request. – 2017-01-24
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    Depending on context, "smooth" can mean "differentiable", but it more usually means "infinitely differentiable". It is possible for $f$ to be infinitely differentiable and for the integral to converge - indeed, for the integral to converge to $0$. It is easier, however, to start by finding an example where $f$ is merely continuous. – 2017-01-25
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    @PatrickStevens If one is thinking "PDFs" and "gaussian densities", then $C^\infty$ examples are actually the first ones that come to mind. If one is thinking "spikes", then indeed, continuous, nonsmooth, examples are on the forefront. – 2017-01-25
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    Ops, what I meant was: if $f(x)$ is a differentiable function over $[a, +\infty)$ and satisfies 1. 2. and 3. than $\int_{a}^{+\infty}f(x)dx\notin\mathbb{R}$ – 2017-01-25

1 Answers 1

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You're entirely correct, assuming by "integral" you mean "Lebesgue integral" for your part 4.

$\int_1^{\infty} f(x) dx$ is the integral of the indicator function of a null set (indeed, all countable subsets of $\mathbb{R}$ have measure $0$), so its value is $0$.

As user251257 points out, it's actually true in the sense of Riemann too. You can easily show that $\int_a^m f = 0$ for all $m$ - as you've pointed out already, $f$ on $[a,m]$ is just the $0$ function with finitely many discontinuities - and so $\int_a^{\infty} f = 0$ too.

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    I have not studied Lebesgue integral yet. I only know Riemann integration, but thank you for your help. I really appreciate it. – 2017-01-24