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Is my proof correct?

Proof: Let $\varepsilon>0$ and consider only refinement of the partition $$P_0=\{ a,c-\frac{\varepsilon}{4},c+\frac{\varepsilon}{4},b \}.$$ This way, $$ U(f,P)=\left[c+\frac{\varepsilon}{4}-\left(c-\frac{\varepsilon}{4}\right)\right]=\frac{\varepsilon}{2}<\varepsilon $$ $$ L(f,P)=0. $$ Thus, for all $\varepsilon>0$ $$ U(f,P)=U(f,P)-L(f,P)<\varepsilon $$ $$ \implies U(f,P)=L(f,P)=0. $$ By definition, $ \underline{\int_{a}^{b}}f\le\overline{\int_{a}^{b}}f $. On the other hand $$ \overline{\int_{a}^{b}}f=\inf\limits_{P}\{U(f,P)\}\le U(f,P)=L(f,P)\le \sup\limits_{P}\{L(f,P)\}=\underline{\int_{a}^{b}}f $$ $$ \implies \underline{\int_{a}^{b}}f\ge\overline{\int_{a}^{b}}f $$ Therefore, $f$ is integrable.

Since for all partitions we have $ L(f,P)=0 $, $$ \sup\limits_{P}\{L(f,P)\}=\underline{\int_{a}^{b}}f=\int_{a}^{b}f=0. $$

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No, it is not correct. What you proved was that for each $\varepsilon>0$ there is a partition $P\varepsilon$ (you called it $P$ but I want to stress that it depends on $\varepsilon$) such that $L(f,P\varepsilon)=0$ and that $U(f,P\varepsilon)<\varepsilon$. So, yes, $U(f,P\varepsilon)-L(f,P\varepsilon)<\varepsilon$, but $U(f,P\varepsilon)$ is still greater than $0$; you cannot say that $U(f,P\varepsilon)-L(f,P\varepsilon)=0$.

However, it follows from what you did that$$(\forall\varepsilon>0):\overline{\int_a^b}f(x)\,\mathrm dx-\underline{\int_a^b}f(x)\,\mathrm dx<\varepsilon$$and therefore, yes$$\int_a^bf(x)\,\mathrm dx=0.$$

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    $\begingroup$ I see, I cannot say that $U(f,P_\varepsilon)-L(f,P_\varepsilon)=0$. But I can say that $\sup\{ L(f,P_\varepsilon\}=0$, right? So, $ \inf\{U(f,P_\varepsilon)\}-\sup\{L(f,P_\varepsilon)\}<\varepsilon.$ And the rest follows from this, is that correct? Anyways thanks for your help. $\endgroup$ – Math_Hater Sep 8 '19 at 12:41
  • $\begingroup$ Yes, that would be correct. $\endgroup$ – José Carlos Santos Sep 8 '19 at 12:43
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I'd do it this way:

Darboux integral

Clearly, $\inf f = 0$ on all of the subintervals of $[a,b]$, which means that $$\underline{\int} f = 0$$ It's also clear that $\sup f = 0$ on all of the subintervals of $[a,b]$ which does not contain $c$, and $\sup f = 1$ otherwise. If $\delta$ is the norm of the partition, then we have that $$0\leqslant U \leqslant 2\delta$$ So we can construct a sequence of partitions for which the norm of the partitions will go to zero, i.e. $\delta_n \to 0$, so for their upper Darboux sum, we will have that $$0 \leqslant U_n \leqslant 2\delta_n$$ Which means that the upper Darboux integral is $0$ as well: $$\overline{\int}f=0$$ So by the definition of the Darboux integrability, $f$ is integrable, and it's integral is $0$.

Riemann integral

Let $\varepsilon > 0$ be given. Let $\delta = \frac{\varepsilon}{4}$. If the partition is finer than $\delta$, then we have that the Riemann sum $S$ is $$0 \leqslant S \leqslant 2 \delta $$ Which means that $$|S-0|\leqslant |2 \delta|=\frac{\varepsilon}{2}<\varepsilon$$ Hence $f$ is Riemann integrable, and it's integral is $0$.

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