Problem is the following:
$\displaystyle\qquad f(x) = \begin{cases} \sin(1/x) & \text{if } x \text{ is irrational} \\ 0 & \text{otherwise } \end{cases}$
Prove that $f(x)$ is not Riemann integrable on $[0,1]$.
My approach:
I can show that for a sub-interval $I_{1} = [2/\pi,1]$,
$\displaystyle \qquad U(I_{1},f)-L(I_{1},f) > \sin(1)\ (1-\frac{\pi}{2}\ )$
But i could not find any property which imply that if $I_{1}$ is not Riemann integrable then interval $A=[0,1]$ is also not Riemann integrable.
HINT: note that for any propositions $A$ and $B$ it holds that $$ (A\implies B)\iff (\lnot B\implies \lnot A) $$ Therefore proving that $$ \int_c^d f(x)\,\mathrm d x\text{ doesn't exists }\implies \int_a^b f(x)\,\mathrm d x\text{ doesn't exists} $$ for $[c,d]\subset [a,b]$, is equivalent to show that $$ \int_a^b f(x)\,\mathrm d x\text{ exists }\implies \int_c^d f(x)\,\mathrm d x\text{ exists } $$ what could be easier to prove. The above strategy is named as proof by contraposition because the contrapositive of the implication $A\implies B$ is $\lnot B\implies \lnot A$.
For your question probably is the same to prove the statement directly or the contrapositive statement, however it is a very useful technique to remember to prove things