Suppose that there is an $M>0$ such that for every $\delta>0$ there are Riemann sums $\sigma_1$ and $\sigma_2$ over a partition P of $[a,b]$ with $||P||<\delta$ such that $|\sigma_1 - \sigma_2| \geq M$. Prove that $f$ is not integrable over [a,b].

Given that for $M>0$, we have
$$|\sigma_1 - \sigma_2| \geq M$$ $$|\sigma_1 -L + L - \sigma_2| \geq M$$ applying the triangle inequality: $$|\sigma_1 -L| + |L - \sigma_2| \geq |\sigma_1 - \sigma_2| \geq M$$ $$|\sigma_1 -L| + |\sigma_2-L| \geq M$$

It shows that there exists at least one of the quantity unbounded (let $\sigma_1$ be considered) It follows that for at least one $\epsilon$, we have $$|\sigma_1 -L|>\epsilon$$ This contradict the definition of $L$ defined as the Riemann integral of $f$ over $[a,b]$, $L=\int_{a}^{b} f(x)\, dx$ as the unique Riemann integral of $f$, where $\forall$P of $[a,b]$: $$\left|\sigma_1 - \int_{a}^{b} f(x)\, dx\right|<\epsilon$$ Therefore $f$ is not integrable over $[a,b]$.

I would like to know if my proof is correctly done but also if there is anything to be added and or to be rectified. Much appreciated.

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    $\begingroup$ Please explain what $L$ is. It comes out of nowhere. $\endgroup$ – zhw. Jul 26 '17 at 16:45
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    $\begingroup$ This is the contrapositive of the question you asked yesterday. $\endgroup$ – MathematicsStudent1122 Jul 26 '17 at 16:53
  • $\begingroup$ Yes it is. checking if I understood your proof you showed yesterday. thx again. $\endgroup$ – gegu Jul 26 '17 at 21:27

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