Let $f(x)= x$ if $x$ is irrational and $0$ if $x$ is rational.
Then $f$ is not integrable because for any partition $P$ of $[0,1]$, the lower sum $L(f,P)=0$, because every sub interval of $P$ contains rationals. Further note that $U(f,P)\geq \int^{1}_{0}x dx = \frac{1}{2}$. Hence for $\epsilon= \frac{1}{2}$, for any partition $P$ of $[a,b]$, we have $U(f,P)-L(f,P)\geq \frac{1}{2}$. Hence $f$ is not integrable. Is my reasoning correct?