closed interval proof on integrability reformulation

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You have a function $f:\mathbb{R}\to \mathbb{R}$ that is defined by $f(x) = 1$ if $x\in \mathbb{Q}$, and $-\sqrt{3}$ if $x\not\in \mathbb{Q}$.

Using the Integrability Reformulation definition, how would I prove $f(x)=-1$ is not integrable on $[-5,-1]$?


My understanding: Is that I must choose some epsilon and make a partition within $[-5,-1]$ and calculate the upper and lower sums, then subtract the upper and lower sums, but I can't seem to make it work and choose an appropriate epsilon. My textbook is pretty scarce on this topic with no good examples. Thank You in advance.

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It doesn't matter what partition you define, every interval in the partition will always have a rational number and an irrational number. So the supremum in each interval in the partition is always $1$ while the infimum is always $-\sqrt{3}$. What can you conclude about the upper and lower sums?