Prove not integrable. Is this correct?

[BustedBreaks]

Let f:[0,1] be defined as f(x)= 0 for x rational, f(x)=x for x irrational

Show f is not integrable

m=inf(f(x) on [Xi-1, Xi])
M=sup(f(x) on [Xi-1, Xi])

Okay so my argument goes like this:
I need to show that the Upper integral of f does not equal the lower integral of f

Because rationals and irrationals are dense in R for any interval [a,b] of f,

m=0
M=x

therefore, for the partition X0=0, X1=1 the Lower Darboux sum = 0 and the Upper Darboux sum = x

therefore the sup{Lower Darboux sum}=0
and the inf{Upper Darboux sum} is not equal to 0, (it is the smallest irrational number greater than 0 technically right? although there is no smallest irrational greater than 0, right?)

therefore the Upper integral of f does not equal the lower integral of f

and therefore f is not integrable.

I think this is right, but I just want to make sure I didn't miss anything.

Thanks!

 

[Office_Shredder]

and the inf{Upper Darboux sum} is not equal to 0, (it is the smallest irrational number greater than 0 technically right? although there is no smallest irrational greater than 0, right?)

For your upper sum you said that M on an interval has to be greater than the smallest irrational inside the interval. When you break up [0,1] into tiny pieces you're essentially integrating the function x, so you expect to get as an upper sum 1/2

All you have to prove is that the upper sum can't give you 0; saying it has to be larger than the smallest irrational number greater than 0 is not good enough (since as yo usaid there is no such number, so that has no meaning)

 

[BustedBreaks]

For your upper sum you said that M on an interval has to be greater than the smallest irrational inside the interval. When you break up [0,1] into tiny pieces you're essentially integrating the function x, so you expect to get as an upper sum 1/2

All you have to prove is that the upper sum can't give you 0; saying it has to be larger than the smallest irrational number greater than 0 is not good enough (since as yo usaid there is no such number, so that has no meaning)

Well, what I was trying to say was that the upper sum cannot equal 0 because for any interval [a,b] M will not be 0 because there is always an irrational greater than 0 no matter how small the partition [Xo, X1] gets. Because M will never be 0 the inf of the Upper Darboux Sums will never be 0 and thus the upper integral will never equal the lower integral, even though I haven't defined what the upper actually is...

I wasn't trying to say anything the smallest irrational number greater than 0, just that it couldn't be equal to 0 because there would always be an irrational greater than 0 making M on that interval greater than 0.