Given the following definition of the Riemann upper integral:
$\bar{\int_{a}^{b}}f =\inf \{ U(f;P): \text{ P is a partition of [a,b] } \}$
Where $U(f;P)=\sum\limits_{i=1}^{n} M_{i}(f)\Delta x_i$ and $M_{i}(f)=\sup\{f(y): x \in [x_{i-1}, x_i] \}$.
I want to show that this definition is equivalent:
$\bar{\int_{a}^{b}}f =\inf \{ U^o(f;P): \text{ P is a partition of [a,b] } \}$
Where $U^o(f;P)=\sum\limits_{i=1}^{n} M_{i}^{o}(f)\Delta x_i$ and $M_{i}^{o}(f)=\sup\{f(y): x_{i-1}
I noticed that in my textbook the discussion of Riemann integrals is the characterized by picking the supremum of an open interval whereas in class we have discussed it in terms of supremum's of a closed interval. We were asked as a non-homework exercise to prove they are equivalent. Our hint was to let $\alpha = \inf\{U^o(f:P)\}$ and show that $\alpha=\inf\{U(f;P)\}$.
I don't see how this proof is straight forward because my thinking is $\sup\{[a,b]\} \geq \sup\{(a,b)\}$.