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Let's say $f:[-1,1]\to\mathbb R$ is a bounded function. Let's say that $f$ is Riemann integrable over $[a,b]$ for every $a$ in $]-1,0[$ and every $b$ in $]0,1[$. Can we conclude that $f$ is Riemann integrable over $[-1,1]$?

I was first looking for a counter example. I see that the critical points are $-1,$ $1,$ and $0.$ I was looking for something with more than one discontinuity point because then it's not Riemann integrable.

I don't really see a counter example can someone maybe give me a hint.

Edit: Because of the information I get in the reaction I'm no longer looking for a counter example but I'm trying to prove that $f$ is Riemann integrable over [-1,1]. Now I tried to prove that the lower sum equals the upper sum, because that's a basic definition for Riemann integrability but I'm a little bit confused because we don't have a concrete function. Maybe someone can suggest another way for a proof?

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    $\begingroup$ This is easy to answer if you know the characterization that a bounded function is Riemann integrable if and only if the set of points where it is discontinuous has measure zero. Adding three more points of possible discontinuity (-1, 0, and 1) won't affect this. $\endgroup$
    – user169852
    Aug 21, 2020 at 19:44
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    $\begingroup$ No, I'm saying that $f$ is Riemann integrable on every $[a,b]$ with $-1 < a < 0 < b < 1$, which means that $f$ is continuous at almost every point in $(-1, 0) \cup (0, 1)$ (i.e. everywhere in these intervals except possibly a set of measure zero). Therefore $f$ is continuous at almost every point in $[-1,1]$ (because if you add three points to a set of measure zero, the result still has measure zero), and therefore $f$ is Riemann integrable on $[-1,1]$. $\endgroup$
    – user169852
    Aug 21, 2020 at 19:51
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    $\begingroup$ Yes, a bounded function is Riemann integrable if and only if it is almost everywhere continuous. A set of measure zero is a set that can be covered by countably many intervals with arbitrarily small positive length. More formally definition here: en.wikipedia.org/wiki/Null_set#Definition $\endgroup$
    – user169852
    Aug 21, 2020 at 19:55
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    $\begingroup$ I'm guessing you don't have that theorem available, so an alternate solution would be use the Riemann sum definition, with a partition of $[-1,1]$ that puts the points -1, 0, and 1 in very narrow rectangles, say $[-1,-1+\epsilon]$, $[-\epsilon, \epsilon]$, and $[1-\epsilon, 1]$. $\endgroup$
    – user169852
    Aug 21, 2020 at 19:58
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    $\begingroup$ @Bungo, no need for $[-\epsilon,\epsilon]$. $\endgroup$
    – Philipp
    Aug 21, 2020 at 23:49

1 Answer 1

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We know that $f:[-1,1]\to \mathbb{R}$ is bounded and Riemann-integrable on $[a,b]$ where $-1<a<0<b<1$.

$f$ is bounded on $[-1,1]$ so there must exist the following supremum und infimum $$M_a=M_b:=\sup\{f(x)~|~x\in [-1,1]\},$$ $$m_a=m_b:=\inf\{f(x)~|~x\in [-1,1]\}.$$

Let be $\epsilon>0$ arbitrary chosen. We now have to choose $a$ and $b$ such that they are sufficiently close to $-1$ and $1$ respectively. Let be $a+1<\frac{\epsilon}{3(M_a-m_a)}$ and $1-b<\frac{\epsilon}{3(M_b-m_b)}$. Consider the equidistant partition $P_n$ of $[a,b]$, $$P_n:=\{a, a+\frac{b-a}{n}, a+2\frac{b-a}{n}, \cdots,a+(n-1)\frac{b-a}{n},b\}.$$

Let be $$M_i:=\sup\{f(x)~|~x\in [a+i\frac{b-a}{n},a+(i+1)\frac{b-a}{n}]\}$$ and $$m_i:=\inf\{f(x)~|~x\in [a+i\frac{b-a}{n},a+(i+1)\frac{b-a}{n}]\}.$$

Then we know that for an arbitrary small $\frac{\epsilon}{3}>0$ there exists a $n_0$ such that every partition $P_n$ with a higher refinement than $P_{n_0}$ satisfies $\sum\limits_{i=1}^{n}(M_i-m_i)\frac{b-a}{n}<\frac{\epsilon}{3}$ (This is the definition of the Riemann-integrability). Now we extend our interval $[a,b]$ to $[-1,1]$ such that our partition $P_n$ becomes $$P_n:=\{-1,a, a+\frac{b-a}{n}, a+2\frac{b-a}{n}, \cdots,a+(n-1)\frac{b-a}{n},b,1\}.$$ Now the sum of the upper and lower Darboux-sums gets two more summands: $$(M_a-m_a)\frac{\epsilon}{3(M_a-m_a)}+\sum\limits_{i=1}^{n}(M_i-m_i)\frac{b-a}{n}+(M_b-m_b)\frac{\epsilon}{3(M_b-m_b)}<\epsilon.$$ Hence, $f$ is Riemann-integrable on $[-1,1]$.

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  • $\begingroup$ How you become a+1< and 1-b< at the beginning? i can't see why? $\endgroup$
    – questmath
    Aug 22, 2020 at 6:27
  • $\begingroup$ @mathmath, we must choose $a$ and $b$ sufficiently close to the bounds $-1$ and $1$, respectively. Hence, we set $|a-(-1)|=a+1<\epsilon'$ and $|b-1|=1-b<\epsilon'$ for a "very small" $\epsilon' >0$. To arrive at the desired result, UpperSum - LowerSum <$\epsilon$, we set $\epsilon':=\frac{\epsilon}{3(M_a-m_a)}$. $\endgroup$
    – Philipp
    Aug 22, 2020 at 12:54
  • $\begingroup$ @mathmath, it's basically what Bungo said in his comment when he gave the advise to put $-1$ and $1$ in very narrow rectangles. $\endgroup$
    – Philipp
    Aug 22, 2020 at 13:11
  • $\begingroup$ i can't understand why you take a maximum and minimum but you write two times the same Ma and Mb are defined the same way? $\endgroup$
    – questmath
    Aug 23, 2020 at 12:21
  • $\begingroup$ @mathmath, yes that is correct. $M_a$ and $M_b$ are the same (so are $m_a$ and $m_b$). By using those indices I just wanted to stress that the two additional summands in the sum of the upper and lower Darboux sums come from $a$ and $b$ respectively. Maybe this was a bit confusing. I will edit my answer regarding this issue. $\endgroup$
    – Philipp
    Aug 23, 2020 at 13:39

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