Suppose that $f:[0:1] \rightarrow \mathbb{R} $ is a bounded function with upper and lower integrals: $$\int_{\underline{0}}^{1}f=0 \ \ \textrm{and } \int_{0}^{\overline{1}}f=1$$
(a) Prove that for every $\epsilon>0 $ there exists a subdivision $\sigma$ of $[0,1]$ such that the difference between the upper and lower sums satisfies $1\le U(\sigma ,f)-L(\sigma ,f)<1+\epsilon$
(b) Does there have to be a subdivision such that $U(\sigma ,f)-L(\sigma ,f)=1?$ Either prove it or find a counterexample.
For part (a), I did as follows:
Let $\sigma$ be the subdivision $\sigma =[a_0 =0\le a_1 \le a_2 ... < a_n=1]$. Since $f$ is bounded, $M_i$ and $m_i$ exist and, thus, $U(\sigma,f)$ and $L(\sigma,f)$ exist.
$\int_{\underline{0}}^{1}f=sup\{ L(\sigma,f), \textrm{all subdivisions } \ \sigma \ \textrm{of} \ [0,1] \}=0$
$\int_{0}^{\overline{1}}f=inf\{ U(\sigma,f), \textrm{all subdivisions } \ \sigma \ \textrm{of} \ [0,1] \}=1$
$m(1-0)<L(\sigma,f)\le U(\sigma,f)<M(1-0)$.
$m<L(\sigma,f)\le U(\sigma,f)<M$
We know that,
$\int_{0}^{\overline{1}}f\le U(\sigma,f)$
Choosing $\epsilon>0$ such that $\int_{0}^{\overline{1}}f=infU(\sigma,f)+\epsilon >U(\sigma,f)$
Thus, $U(\sigma,f)<1+\epsilon$ since $U(\sigma,f)>L(\sigma,f)$ and $U(\sigma,f)<1+\epsilon$. We can say $U(\sigma ,f)-L(\sigma ,f)<1+\epsilon.$ We also know that $\int_{\underline{0}}^{1}f\ge L(\sigma,f)$ and $\int_{0}^{\overline{1}}f\le U(\sigma,f).$
Therefore, $\int_{\underline{0}}^{1}f - \int_{0}^{\overline{1}}f \le U(\sigma ,f)-L(\sigma ,f)$ so $1-0=1\le U(\sigma ,f)-L(\sigma ,f)<1+ \epsilon$
Have I done part (a) correctly? Also, how should I approach part (b) as I am stuck on it?