# Show that f is not integrable

1. Jan 10, 2009

### snipez90

1. The problem statement, all variables and given/known data
Let f(x) = 0, if x is rational; 2 + x, if x is irrational, for all x in [0,1]. Show that f is not integrable

2. Relevant equations
density of rationals/irrationals, equation for L(f,P) and U(f,P), Darboux Integrability criterion

3. The attempt at a solution
L(f,P) = 0 because on any subinterval formed from two consecutive points in the partition of [0,1], there exists a rational x.

For U(f,P), I was thinking that I could just pick any irrational x so that sup{ f(x): x is an element of the subinterval from two consecutive points in the partition of [0,1] } = 2 + x. Then note that since $$2+x \geq 2$$, $$U(f,P) \geq 2(b-a) > 0$$ so that

$$sup {L(f,P)} \neq inf {U(f,P)}$$

I left a few steps in demonstrating U(f,P) > 0 but is this a good approach?

2. Jan 10, 2009

### Dick

Sort of. 2+x>=2. x<=1. Can you show any U(f,P)>=2 and any L(f,P)<=1?

3. Jan 10, 2009

### snipez90

Hmm I was really lax about the bounds. U(f,P) >= 2(b-a) = 2. L(f,P) = 0, no matter what, I think, so L(f,P) is certainly <= 1.

As for U(f,P), could I just say that 2+x >= 2 for any x in [0,1] so that U(f,P) >= 2*SUMMATION{bunch of points that cancel out to give b -a}?

4. Jan 10, 2009

### Dick

Sure U(f,P)>=2, but why do you think L(f,P)=0? Take the partition P of the intervals [0,1/2] and [1/2,1]. Then L(f,P)=1/4, I think.

5. Jan 10, 2009

### snipez90

Sorry, here's my reasoning. The definition of L(f,P) is the summation of a bunch of products, each of which is of the form m*(d-c), where d and c are consecutive points in the partition and c < d. The quantity m is equal to inf{f(x) : c <= x <= d}. But assuming the density of the rationals, the quantity m will always be 0. Therefore, the summation of these products will yield 0 as well.

6. Jan 10, 2009

### Dick

Oooops. Right. I misread the problem. Somehow I read it as f(x)=x for x rational instead of f(x)=0. So, yes, L(f,P)=0, U(f,P)>=2.

7. Jan 10, 2009

### snipez90

Haha ok. Actually when you suggested x <= 1 I thought to myself did I put x instead of 0? Thanks for your help though. Especially in noting that it's probably a good idea to use the best bounds possible (I sometimes forget that b and a are given endpoints and leave b-a as b-a).