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I'm trying to figure out a lecture example given on our Analysis course. We are currently going through Riemann integrals.

Let $g:[0,1] \to R, g(x) = 1$ when $x \in [0, \frac{1}{2}]$ and $g(x) = 2$ when $x \in ]\frac{1}{2},1]$.

Is g integrable?

The example goes on to prove that g is indeed integrable by choosing $P_n = \{0, \frac{1}{2}, \frac{1}{2} + \frac{1}{n}, 1\}$ as the partition.

I don't understand why simply $P = \{0, \frac{1}{2}, 1\}$ isn't enough to prove that the lower and upper Darboux integrals are the same, thus $g$ is integrable.

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Typically the endpoints are included in defining the lower and upper sums, so the lower sum for your partition is $1$ and the upper sum is the correct value of $\frac{3}{2}$. On the other hand, by adding $\frac{1}{2}+\frac{1}{n}$, the discrepancy caused by that pesky endpoint shrinks to $\frac{1}{n}$, which goes to $0$.

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    Yeah, seems like it and I suppose all the supporting lemmas etc won't change either. Good question @randomguy :-)2011-01-28