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$$ f(x) = \begin{cases} -2, & \text{if }x < 0 \\ 1, & \text{if }x > 0\\ 0, & \text{if }x = 0 \end{cases} $$

Hey guys I need some help showing that this function is integrable on the closed interval $[-1,2]$.

So far my idea has been to show $$U(f,P)-L(f,P) < \epsilon$$ for some $\epsilon>0$.

The only problem is the point $(0,0)$ on the function.

I don't understand how to handle that.

Can I just say that $U(f,P)$ for some partition will equal to $3$ and then find a partition $P$ for which $$3-L(f,P)<\epsilon?$$

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  • $\begingroup$ If you make the interval(s) containing 0 sufficiently narrow, you can make the difference between upper an lower sums as small as you want, namely at most $|1-(-2)|=3$ times this interval width. $\endgroup$ – Henning Makholm Nov 18 '15 at 23:28
  • $\begingroup$ Hint: look at partitions like $\{[-1,-\epsilon],[-\epsilon,\epsilon],[\epsilon,2]\}$ for smaller and smaller (positive) $\epsilon$. $\endgroup$ – Arthur Nov 18 '15 at 23:29
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Let $\sigma$ be a partition of $[-1,2]$. Taking Arthur's hint, have the partition be of the form, $$[-1,-\delta],[-\delta,\delta],[\delta,2]$$ for appropriately small $\delta >0$.

The supremum of $f(x)$ is -2 on the first subinterval, $1$ on the second subinterval, and $1$ on the third subinterval (verify it!). Thus the upper sum is given by $$U_{f,\sigma}=(1-\delta)\cdot -2 + 2\delta \cdot 1 + (2-\delta)\cdot 1=3\delta$$ The infimum of $f$ on the first interval is $-2$, as well as $-2$ on the second subinterval, and $1$ on the last (verify). Thus the lowers sum is given by $$L_{f,\sigma}=(1-\delta)(-2)+2\delta \cdot (-2) +(2-\delta)\cdot 1=-3\delta$$ Thus, $U_{f,\sigma}-L_{f,\sigma}=6\delta$ and it should be clear how small to make $\delta$.

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  • $\begingroup$ So after you get 2$\delta$ we need a $\delta$ < $\epsilon$ /2. We could choose $\epsilon$ /3 and get Uf-Lf=$\epsilon$ - $\epsilon$ /3 which finishes the proof of 2/3 $\epsilon$ < $\epsilon$. Would that be correct? $\endgroup$ – jeff Nov 19 '15 at 0:54
  • $\begingroup$ Yes, you got it. $\endgroup$ – Nap D. Lover Nov 19 '15 at 1:03
  • $\begingroup$ Just one question though. Why isn't the infimum on the interval [-$\delta$,$\delta$] set as -2? $\endgroup$ – jeff Nov 19 '15 at 1:07
  • $\begingroup$ Because i messed up haha. This will affect the bound $\endgroup$ – Nap D. Lover Nov 19 '15 at 1:48
  • $\begingroup$ @jeff I have edited my answer with this fixed now. $\endgroup$ – Nap D. Lover Nov 19 '15 at 1:53
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Notice that for any $x$ at which $f$ is discontinuous, $x$ can be contained in an interval as small as you please. Make sure to include this small interval in your partition.

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