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"Let $f(x)=1$ if $x=1,1/2,1/3,1/4,...$ and $f(x)=0$ elsewhere. Prove that $f$ is integrable on $[0,1]$. What is the value of that integral?"

I'm guessing the value to be $0$, intuitively. I know the lower integral is zero. But I can't prove the same about the upper integral because I can't find the appropriate partition. Part of me says its not integrable but not sure about that.

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  • $\begingroup$ If you are allowed to use $\int_0^1f(x)dx=\lim_{x\rightarrow0^{+}}\int_x^1f(x)dx$ then you only have to deal with a finite number of points of discontinuity. $\endgroup$ – Gregory Grant May 16 '15 at 15:17
  • $\begingroup$ @GregoryGrant Actually I'm not allowed to use that. I'm only given the whole section about upper lower sum and integral tho. $\endgroup$ – BMS May 16 '15 at 15:23
  • $\begingroup$ I posted an idea, let me know if that's acceptable. If not I'll try to help modify it. $\endgroup$ – Gregory Grant May 16 '15 at 15:27
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Let $k\in\mathbb N$ and $\epsilon=\frac1{k}>0$. Then all but finitely many of the points of discontinuity are in $[0,\epsilon]$. All but $k-1$ points to be exact. Suppose $x_1,\dots,x_{k-1}$ are the points of discontinuity outside that interval (in order). Note that $x_{k-1}=1$. Now take the partition $$0,\ \epsilon,\ x_1-\frac{\epsilon}{k},\ x_1+\frac{\epsilon}{k},\ \dots,\ x_{k-2}-\frac{\epsilon}{k},\ x_{k-2}+\frac{\epsilon}{k},\ 1-\frac{\epsilon}{k},\ 1$$ The upper sum is equal to $$\epsilon + \left((k-2)\frac{2\epsilon}{k}\right)+\frac{\epsilon}{k}=\frac{1}{k}+\frac{2k-3}{k^2}$$ This goes to zero as $k\rightarrow\infty$. And the lower sum is zero. So the difference of upper sum and lower sum for this sequence of partitions goes to zero. So the function is Riemann integrable. And since the sums converge to zero, $\int_0^1f=0$.

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  • $\begingroup$ Nice idea. Did you get the partition by looking at the graph? I'm still trying to grasp the "upper sum is equal to" part. How can I be sure the supremum of each subinterval is equal to 1? Or it isn't? $\endgroup$ – BMS May 16 '15 at 15:51
  • $\begingroup$ Yes, the idea is an arbitrarily small interval $[0,\epsilon]$ contains all but finitely many of the problematic points. You then just have to squeeze down tightly around those finitely many points. $\endgroup$ – Gregory Grant May 16 '15 at 15:56
  • $\begingroup$ To get the upper sum, consider it by cases. First consider $[0,\epsilon]$ and $[1-\epsilon/k,1]$, the first and last intervals in the partition. The first contributes $\epsilon$ to the upper sum, the second contributes $\epsilon/k$. Now, the ones in between are either $[x_i+\epsilon/k,x_{i+1}-\epsilon/k]$ or $[x_i-\epsilon/k,x_i+\epsilon/k]$. The first contribute zero to the upper sum ($f$ is zero on those intervals) and the second, contributes $2\epsilon/k$. And there are exactly $k-2$ of them. Does that help? $\endgroup$ – Gregory Grant May 16 '15 at 15:57
  • $\begingroup$ I rewrote it without the sigma notation for the sum, because maybe that was just making it more confusing than necessary. It's just $k-2$ copies of $2\epsilon/2$. $\endgroup$ – Gregory Grant May 16 '15 at 15:59
  • $\begingroup$ Oh, so the interval length of the first one is $x_{i+1}-x_i-2\epsilon/k < x_{i+1}-x_i$ and thus contributes zero, right? And also, the supremum in each of the other intervals is $1$ so that each contributes $2\epsilon/k$? $\endgroup$ – BMS May 16 '15 at 16:12
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Use can use the fact that That the function is bounded on a closed interval to get that it is Riemann integrable and also Lebesgue integrable. The Lebesgue integral is $0$ and therefore the same for Riemann.

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