Examples of Non-Riemann Integrability

**ragnar** · June 23rd 2010, 10:41 PM

I was wondering if people can give me "nice" examples of non-Riemann integrable functions. I know the one about the rationals and irrationals, so-called indicator function (and called something else by a lot of other people), but I was hoping for something a little more natural. I'm trying to see whether $f^{2} \in \mathscr{R}$ entails that $f \in \mathscr{R}$ , and I thought that if I had a few examples of non-integrable functions it might help, but there are precious few to be found by a basic Google search.

**Ackbeet** · June 24th 2010, 01:47 AM

Well, we know that a function is Riemann integrable if and only if its set of discontinuities has measure zero. So, you're going to have to have a function that has loads of discontinuities in a very small space. I don't know of any other examples other than the characteristic on the rationals (or irrationals, it doesn't matter which). I'm sure you could construct many such functions, but I think they would all have to be a similarly pathological function to the characteristic on the rationals.

$f^{2}$ being Riemann integrable does not imply that $f$ is Riemann integrable. Counter example:

Let $f:[0,1]\to[-1,1]$ be defined by

$f(x)=\begin{cases}1\quad x\in\mathbb{Q}\cap[0,1]\\<br /> -1\quad x\in\mathbb{Q}^{c}\cap[0,1]\end{cases}.$

Then, clearly, $f$ suffers from the same problem that the characteristic function does: its set of discontinuities has measure 1. However, $f^{2}(x)=1$ for all $x\in[0,1]$ . That function is obviously Riemann integrable.

**gosualite** · June 24th 2010, 07:50 AM

Try an unbounded function:

Take $f\colon [0,1] \to \mathbb{R}$ giving $f(x) = \begin{cases} 1/\sqrt{x}, & x > 0\\ 0, & x = 0.\end{cases}$

which is Riemann integrable on all proper closed subintervals of [0,1], but is not Riemann integrable on [0,1] itself. I chose $1/\sqrt{x}$ because the area under this graph from 0 to 1 is finite and can still be computed using elementary calculus techniques, yet it's not Riemann integrable by the definition.

**chiph588@** · June 24th 2010, 08:51 AM

Originally Posted by gosualite

Try an unbounded function:

Take $f\colon [0,1] \to \mathbb{R}$ giving $f(x) = \begin{cases} 1/\sqrt{x}, & x > 0\\ 0, & x = 0.\end{cases}$

which is Riemann integrable on all proper closed subintervals of [0,1], but is not Riemann integrable on [0,1] itself. I chose $1/\sqrt{x}$ because the area under this graph from 0 to 1 is finite and can still be computed using elementary calculus techniques, yet it's not Riemann integrable by the definition.

Why is exactly is $f(x)$ not Riemann integrable here?

**Ackbeet** · June 24th 2010, 04:17 PM

You might check and see if your library has this book (or you can just buy it - it's not expensive). You might find something in there.

**Jose27** · June 24th 2010, 04:24 PM

Originally Posted by chiph588@

Why is exactly is $f(x)$ not Riemann integrable here?

Because, by definition, a Riemann integrable function must be bounded. Integrals of the type gosualite gave are called improper Riemann integrable.

**Ackbeet** · June 24th 2010, 04:24 PM

I agree with chiph588: the exhibited function is actually Riemann integrable, and even finite. To find a function along those lines that has infinite area, you have to go to the other side of 1 in the exponent:

$f(x)=1/x^{2}$ has infinite integral on that interval.

But that's not the same thing as being Riemann integrable or not. The set of discontinuities of $1/x^{2}$ still has measure zero, thus implying that the function is Riemann integrable. The integral just has an infinite value. The characteristic function on the rationals is a bona fide non-Riemann-integrable function, because it's everywhere discontinuous.

The problem in finding non-Riemann-integrable functions is that you have to pick up enough points of discontinuity on the real axis to get positive measure. That means irrationals, and that means you're going to have to have a function that oscillates so wildly that it looks something like the characteristic function of the rationals.

**Ackbeet** · June 24th 2010, 04:28 PM

Reply to Jose27:

Perhaps you're right. I think some authors just lump all the Riemann integrable functions together. In my analysis course, it was Riemann integrable if and only if set of discontinuities has measure zero, regardless of boundedness.

**Jose27** · June 24th 2010, 04:58 PM

Originally Posted by Ackbeet

Reply to Jose27:

Perhaps you're right. I think some authors just lump all the Riemann integrable functions together. In my analysis course, it was Riemann integrable if and only if set of discontinuities has measure zero, regardless of boundedness.

The main problem I have with that definition is that it doesn't coincide with the partition or step-function-aprox. definition of the Riemann integral (the problem being taking a supremum or infimum), and although one can arrive at Lebesgue's criterion and from there drop the boundedness condition, it is really an abuse of notation to call them Riemann integrable, since by definition they're not. Though it's not like mathematics isn't full of notation abuse.

**Ackbeet** · June 24th 2010, 05:14 PM

Perhaps you're right. I, for one, though, would like the integral exhibited in post #3 to be Riemann-integrable, since it does have a finite area in that interval. In one sense, your way is tidier, but in another, mine is. I'm curious how the improper Riemann integral is defined - I'm not sure I've run across it in many analysis books. Could you point me to a reference, please?

**Ackbeet** · June 24th 2010, 05:19 PM

Wait, I see it. They're defined as limits of Riemann integrals. Got it.

**Jose27** · June 24th 2010, 05:23 PM

Originally Posted by Ackbeet

Perhaps you're right. I, for one, though, would like the integral exhibited in post #3 to be Riemann-integrable, since it does have a finite area in that interval. In one sense, your way is tidier, but in another, mine is. I'm curious how the improper Riemann integral is defined - I'm not sure I've run across it in many analysis books. Could you point me to a reference, please?

They're treated as exercises in Spivak's Calculus (chapter 14). They're essentially of two types.

Another thing: your definition, as given, still requires some method of approximation (which is it?) since otherwise $\frac{1}{x}$ would be integrable in any closed interval containing 0.

**Ackbeet** · June 24th 2010, 05:30 PM

No, I don't think 1/x would be integrable in any interval containing 0, because the limits which you use to define the integral would not exist. The limits have to exist in order for the improper Riemann integral to be well-defined.

**Jose27** · June 24th 2010, 05:37 PM

Originally Posted by Ackbeet

No, I don't think 1/x would be integrable in any interval containing 0, because the limits which you use to define the integral would not exist. The limits have to exist in order for the improper Riemann integral to be well-defined.

Sure, it is not improper Riemann integrable, but what I'm asking is how do you arrive at the non-integrability of such a function with your definition, since its set of discontinuities clearly has measure 0 (and it's not improper R. or Lebesgue integrable).

**Ackbeet** · June 24th 2010, 05:42 PM

Well, with my definition, I would say that a function can be Riemann integrable, and yet still have an infinite value.

You know, it just occurred to me that I might be digging myself in a hole here with wrong definitions. I could easily be wrong about this whole darn thing. Let me check my analysis book tomorrow (it's at work), and I'll see whether "bounded" is in the Lebesgue criteria or not. Then I'll get back to you.

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