Greatest Integer Function $ [x^2] $ : Riemann Integration Question How to show $$ [x^2]  $$ is Riemann Integrable in [0,2] ?
I will explain how I proceeded my doubt is with greatest integer function part, 
after splitting into 3 intervals 
$$[0,1],[{1,\sqrt 2 }], [{\sqrt 2,\sqrt 3 }], [{\sqrt 3,2] }$$  
In the interval $$[0,1]$$ Supremum of function is 1 and Infimum is 0.
In the interval $$[{1,\sqrt 2 }]$$ Supremum of function is 2 and Infimum is 1.
In the interval $$[{\sqrt 2,\sqrt 3 }]$$ Supremum of function is 3 and Infimum is 2.
Now U(P,f) = $$ 1(1-0)+2(\sqrt 2-1)+ 3(\sqrt 3-\sqrt 2)+4(2-\sqrt 3)                 $$
and L(P,f) = $$ 0(1-0)+1(\sqrt 2-1)+ 2(\sqrt 3-\sqrt 2)+3(2-\sqrt 3)                 $$
and they are different values. Please help.
 A: Since no one is helping with the answer, I figured out an answer after referring through various questions in Riemann Integrability.
Since the function $ [x^2]  $ is discontinuous at $1,\sqrt 2, \sqrt 3  $ in the interval [0,2], 
we partition the interval as follows:
x0($=0$), x1 , x2 .... xi, y0($=1$), y1 , y2.... yi,z0 (=$\sqrt 2$),z1 , z2.... zi, k0 (=$\sqrt 3$),k1 , k2.... ki,m0$(=2)$
Now U(P,f)= 0*$\Sigma$xi + 1*(y0-xi) +1*$\Sigma$yi + 2*(z0-yi) + 2*$\Sigma$zi +3*(k0-zi)+ 3*$\Sigma$ki + 4*(m0-ki)   
=$0+1*(\sqrt2-1)+2*(\sqrt3-\sqrt2)+3*(2-\sqrt3)+$1*(y0-xi) +2*(z0-yi) +3*(k0-zi) + 4*(m0-ki)
Take
        $\epsilon$ = 1*(y0-xi) +2*(z0-yi) +3*(k0-zi) + 4*(m0-ki)
Now L(P,f)= 0*$\Sigma$xi + 0*(y0-xi) +1*$\Sigma$yi + 1*(z0-yi) + 2*$\Sigma$zi +2*(k0-zi)+ 3*$\Sigma$ki + 3*(m0-ki)   
=$0+1*(\sqrt2-1)+2*(\sqrt3-\sqrt2)+3*(2-\sqrt3)+$  1*(z0-yi) +  +2*(k0-zi)+  3*(m0-ki)  
Now taking U(P,f) - L(P,f) <  $\epsilon$ 
Hence $ [x^2]  $ is Riemann Integrable in [0,2] 
A: I think the conceptual issue is that you're choosing specific partitions that don't have good behavior and using that to argue that the function is not Riemann integrable.
Let me ask you this: would you also think that $x^2$ is not Riemann integrable on $[0,2]$ since on $[0,1]$ the max is $1$ and min is $0$ and on $[1, 2]$ the max is $4$ and min is $1$? I would hope not.
The heart of what it means for a function to be Riemann integrable is you get these upper and lower sums to be arbitrarily close to each other when you choose the right partition. More often than not, the "right" partition is simply one whose parts are sufficiently small. You can now clearly see why continuity is sufficient for Riemann integrability since continuity guarantees that the difference between the max and min in each part is small when the part is small.
