Let's assume that "f=2" if "x\\in Q"
For any partition "0<x_1<x_2<...x_n<1" the Riemann's upper integral is
"\\lim\\limits_{n\\to\\infty}\\sum\\limits_{k=1}^n(x_k-x_{k-1})\\sup\\limits_{x_{k-1}<x<x_k}f(x)=2\\lim\\limits_{n\\to\\infty}\\sum\\limits_{k=1}^n(x_k-x_{k-1})=2"
because in each interval "(x_{k-1};x_k)" there is a rational number "q" and hence the supremum is 2.
The lower integral is
"\\lim\\limits_{n\\to\\infty}\\sum\\limits_{k=1}^n(x_k-x_{k-1})\\inf\\limits_{x_{k-1}<x<x_k}f(x)=0"
because in each interval "(x_{k-1};x_k)" there is an irrational number "\\alpha" and hence the infimum is 0.
Since upper and lower interals are not equal, the function is not Riemann integrable.
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