Riemann Integrability and sign

Question

Given a function $f: [0,1]\rightarrow \mathbb{R}$ defined as $f(x)= x$ if $x\in \mathbb{Q}$ and $-x$ if $x\notin \mathbb{Q}$. I've been told that given the upper $(U(f,D))$ and lower sums $(L(f,D))$ of $f$ for any partition P, we have that $U(f,D)\geq \int^{1}_{0}x dx$ and $L(f,D)\leq \int^{1}_{0} -x dx$. May someone elaborate on why the $-x$ term appeared in the second one, please? I do not want a geometric argument.

Answer 1

Let $D=\{a_0=0,a_1,\ldots,a_n=1\}$, with $a_0<a_1<\cdots<a_n$ be a partition of $[0,1]$. Then, for each $k\in\{1,2,\ldots,n\}$,$$\sup_{x\in[a_{k-1},a_k]}f(x)=a_k=\operatorname{Id}(a_k)$$and$$\inf_{x\in[a_{k-1},a_k]}f(x)=-a_k=-\operatorname{Id}(a_k).$$But then\begin{align}U(f,D)&=a_1(a_1-a_0)+a_2(a_2-a_1)+\cdots+a_n(a_n-a_{n-1})\\&=U(\operatorname{Id},D)\\&\geqslant\int_0^1x\,\mathrm dx\end{align}and, by a similar argument,$$L(f,D)\leqslant\int_0^1-x\,\mathrm dx.$$