# Let $c\in[a,b]$ and $f:[a,b]\to\mathbb{R}$ defined by $f(c)=1$ and $f(x)=0$ for all $x\neq c$. Prove that $f$ is integrable and the integral is zero.

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 $$L(f,P)=0$$, $$\sup\limits_{P}\{L(f,P)\}=\underline{\int_{a}^{b}}f=\int_{a}^{b}f=0.$$

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$$\int_a^bf(x)\,\mathrm dx=0.$$

• 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. – Math_Hater Sep 8 '19 at 12:41
• Yes, that would be correct. – José Carlos Santos Sep 8 '19 at 12:43

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 $$0$$.