let $f(x)=1-x$ for $0 \le x \le 1$.Prove that $U(P,f) > \frac12$ for any partition $P$ of $[0,1]$
We have $$\int_0^1 f(x) \,dx= \frac12.$$Then $\inf U(P,f)= \frac12$ but how to show that $\frac12$ is not attained in any $U(P,f)$ for any partition $P$ of $[0,1]$
Consider the term $\sum_{i=0}^{n-1} \text{sup}_t f(t) (x_{i+1} - x_i)$. Since $f(x) = 1-x$; sup$_tf(t) = 1 - x_{i}$. Thus: \begin{align} U(f,P) &= \sum_{i=0}^{n-1} 1\cdot(x_{i+1} - x_i) - \sum_{i=0}^{n-1} (x_{i}x_{i+1} - x_i^2) \\ &= 1 - [\sum_{i=0}^{n-1} x_{i}x_{i+1} - \sum_{i=0}^{n} x_i^2 + 1] \\ &= - [ \frac{1}{2}(\sum_{i=0}^{n-1}-(x_i - x_{i+1} )^2 - x_0^2 - x_n^2)] \\ &= \frac{1}{2} + \frac{1}{2}(\sum_{i=0}^{n-1}(x_i - x_{i+1} )^2) >\frac{1}{2} \\ \end{align} (Using $x_0 = 0$ and $x_n = 1$.)