Riemann integrability on $[-1,1]$ 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?
 A: 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 $[-1,1]$.
