I'm quite stuck on how to prove that this function: $$ f(x) = \begin{cases} 1 & x \in [0,{1\over 2}) \\ x - {1\over 2} & x \in [{1 \over 2}, 1] \end{cases} $$ is Riemann integrable. I've tried setting the partition $P_1 = \{0,{1\over 2} - \delta, {1\over 2} + \delta, 2\}$, but that turned out not to work, and I've tried a partition $P_N$ of $2N$ evenly spaced points, but it seems that for $f(x) = x - {1\over 2}$, then $U_{P}$ and $L_P$ will always differ by ${1\over 2N}$ for $N$ intervals, i.e., they will always differ by ${1\over 2}$ so I'm slightly stuck. thanks for any tips.
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$\begingroup$ Notice that the mesh of the partition you have is also getting finer. $\endgroup$– user39431Nov 7, 2013 at 23:16
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1 Answer
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Notice that, a function $f$ is Riemann integrable on $[a,b]$ if
i) it is bounded,
ii) it has a countable number of discontinuities ( or it is discontinuous on a set of measure zero. )
answered Nov 7, 2013 at 23:27
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$\begingroup$ I understand, however, I am trying to prove this Riemann integrability from first principles, and so would like to find a way that involves a choice of partition rather than appeal to another theorem $\endgroup$– ModeratNov 8, 2013 at 2:41