Question

\[f(x) =\begin{cases}0 & x = \text{rational}\\1 & x = \text{irrational}\end{cases}\]
Find \[\int_0^1f(x)dx\]

Answer 1

You want to integrate a non-integrable function? D:

Answer 2

Hmmmm... Why not?

Answer 3

Why it's not integrable?

Answer 4

Yes, why is it not integrable?

Answer 5

Upper and lower sums are never equal.

Answer 6

Are you sure that it is NOT integrable?

Answer 7

Quite... it is frequently used as an example of something which isn't integrable.
Let me show you what I mean... when you divide it into sub-intervals, each sub-interval will inevitably contain both rational and irrational numbers, right?

Answer 8

But can't we select a VERY small interval which contains only irrational numbers?

Answer 9

No, that's the beauty of it, there is always a rational number between any two irrational numbers, and vice versa, that's the density property :D

Answer 10

Then what about this?
"Let c be an irrational number in [0,1] and ϵ > 0. Then there is a natural number n such that 1/n < ϵ. If we choose δ small enough that the interval (c-δ, c+δ) contain no rational numbers with denominator less than n, then it follows that for x in this interval, we have f(x)-f(c)| = |f(x)| ≤ 1/n < ϵ"

Extracted from http://www.math.harvard.edu/~ctm/home/text/class/harvard/112/02/html/home/solns/sol11.pdf

Answer 11

While I don't really see what that proves, but suppose we have two irrational numbers, \(\alpha\) and \(\beta\) (or real numbers, it don't matter ;) )

Then, suppose \(\beta > \alpha\) (one of them has to be bigger than the other of course)

Then, \[\beta - \alpha > 0\]
There must exist a positive integer that is bigger than \(\LARGE \frac1{\beta-\alpha}\) , let that integer be n.
\[\Large n > \frac{1}{\beta - \alpha}\]\[\Large n\beta - n\alpha > 1\]This means there is an integer m between \(n\alpha \) and \(n\beta\) since their difference is more than 1.
\[\Large n\alpha < m < n\beta\]\[\Large \alpha < \frac{m}n<\beta\] and there's your rational number.

Answer 12

"This means there is an integer m between nα and nβ since their difference is more than 1." Why must m be an integer?

Answer 13

If two numbers are more than 1 unit apart, there must be an integer between them.

Answer 14

While I think of how to prove this, just you try to find a counterexample...

now let's see...

Answer 15

okay, a rough sketch... first get rid of the trivial case, when one of them is an integer, obviously, if the smaller number \(\alpha\) is an integer, then \(\alpha + 1\) is the integer between \(\alpha\) and \(\beta\). A similar argument, involving \(\beta - 1\) goes if \(\beta\) is an integer.

Now, every non-integer is between two integers...

Answer 16

So that \(\Large \alpha < \lceil \alpha \rceil\)

Answer 17

But \[\Large 0<\lceil \alpha \rceil - \alpha <1\] since \(\alpha \) is not an integer.

Answer 18

\[\Large \beta - \alpha > 1 > \lceil \alpha \rceil-\alpha \]\[\Large \implies \beta - \alpha > \lceil\alpha\rceil-\alpha\implies \beta> \lceil\alpha\rceil\]
\[\Large \implies \alpha < \lceil\alpha\rceil< \beta\]

Answer 19

Thus there must be an integer between two numbers if their difference is more than 1.

Answer 20

I got the part that "when you divide it into sub-intervals, each sub-interval will inevitably contain both rational and irrational numbers". Please move on.

Answer 21

Okay, well, once you divide the interval into sub-intervals, the upper sum would always be 1, right? Since in each interval, the highest possible value f could take is 1, since each interval has irrationals...

Answer 22

Yes.

Answer 23

On the other hand, the lower sum would always be zero since the lowest possible value f could take is 0, since each interval would have rationals.

Answer 24

Wait... lower sum or lower limit?

Answer 25

lower sum, lower limit may or may not refer to the same thing, I'm not aware :)

Answer 26

Ok...

Answer 27

Well, the upper sum is always 1, the lower sum is always 0, the upper sum and lower sum must always approach each other (so to speak) for the function to be Riemann Integrable...

Answer 28

Why in the pdf does it mention that it is integrable?

Answer 29

Interesting. Lebesgue integral by any chance?

Answer 30

Would you please elaborate it?

Answer 31

I don't know it myself, I just saw it in passing that the Dirichlet function (a function similar to this one) is *not* Riemann Integrable, but it is Lebesgue Integrable.

That's all I know on the matter :>

Answer 32

Alright. Thanks for your help!

Answer 33

Do you know if anyone might know this?

Answer 34

Try the big guys :D

Answer 35

Feel free to tag them :)

Answer 36

You tag them :P

Answer 37

I'd rather not bother them myself XD

Answer 38

Right, You need to do something if you want to learn, so...
@amistre64 @satellite73 @zarkon @hartnn @KingGeorge
The questions are
1) Why is \(f(x) =\begin{cases}0 & x = \text{rational}\\1 & x = \text{irrational}\end{cases}\) integrable over [0,1]?
2) How to find \(\int_0^1f(x)dx\)?
Any help would be highly appreciated!

Answer 39

<fingers crossing>

Answer 40

First off, \(f(x) =\begin{cases}0 & x = \text{rational}\\1 & x = \text{irrational}\end{cases}\) is not Riemann integrable for the reasons that @terenzreignz gave above. However, it IS Lebesgue integrable. This can be seen by noticing that the set \(\mathbb{Q}\cap[0,1]\) is a countable set, and thus has measure 0. So the set \([0,1]\setminus(\mathbb{Q}\cap[0,1])\), which is just the irrational numbers in the interval [0,1], has measure 1. So by the definition of Lebesgue integral on indicator functions, we have that\[\int_0^1f(x) dx=1.\]