Remark 11.1.2 (Deficiencies of the Riemann Integral).label [1] Consider the function
\begin{align*}f(x)=\begin{cases}\frac{1}{\sqrt{x}}&x>0\\ 0&x=0\end{cases}\end{align*}
then $f\nin RI([0,1])$ since $U(f,[0,1])=\infty$. Additionally for $\curl{r_n}_{1}^{\infty}=\Q\cap [0,1]$ consider the family of functions
\begin{align*}f_{k}(x)=\begin{cases}1&x\in \curl{r_1,...,r_k}\\ 0&\text{else}\end{cases}\end{align*}
then $f_{k}\in RI([0,1])$ for all $k\geq 1$ but $f(x)\define\limit{k\to \infty}f_{k}(x)=\begin{cases}1&x\in \Q\cap [0,1]\\ 0&\text{else}\end{cases} \nin RI([0,1])$ [2]
- Riemann Integration does not, in general, behave well with respect to integrating discontinuous function and with respect to taking limitskeyboard_return
- If these remarks are still here then we haven’t done any real content yetkeyboard_return
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