# Why isn't this simpler partition enough?

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.

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 $0$.

• Cool! I got it man! Jan 27, 2011 at 23:53
• I now wonder, if we replaced the definition of the Darboux partitions as the union of open/half-open intervals, instead of all closed intervals, will things change? I expect they won't ( countable discrepancies and all), and If they really don't, I suppose randomguy has a valid proof :-) Jan 28, 2011 at 0:00
• @Moron: I'm pretty sure that things don't change, and to prove it one can basically repeat the trick from randomguy's example to approximate mixed-interval sums with closed-interval sums. Jan 28, 2011 at 0:10
• Yeah, seems like it and I suppose all the supporting lemmas etc won't change either. Good question @randomguy :-) Jan 28, 2011 at 0:14