I have a question about 'choosing partition points' for Riemann integrability.
For example,
Let $f:[0,1]\to\mathbb{R}$ be a function defined by $$f(x)\left\{\begin{array}{cc} 0, & x\in[0,1/2], \\ 1, & x\in(1/2,1]. \end{array}\right.$$ Then, for any given $\varepsilon>0$, choose a partition as $\left\{0,1/2-\varepsilon,1/2+\varepsilon,1\right\}$. I've always done this way that 'avoided discontinuities'.
But, i think a partition $\left\{0,1/2,1\right\}$ deduces the Riemann integrability of $f$ on $[0,1]$ as well.
Is it suitable choice to show the Riemann integrability for $f$?
Give some advice/comment. Thank you!
Well, if you choose $\{0,1/2,1\}$, then you have an upper sum of $1/2$ and a lower sum of $0.$ So, the problem is a little more nuanced then that. You can use $$\{0,1/2,1/2+\epsilon,1\}$$ though. This gives an upper sum of $1/2$, and a lower sum of $1/2-\epsilon$, which is enough to deduce Riemann integrability.