Equivalent definition of Riemann integrals 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 $M_{i}^{o}(f)=\sup\{f(y): x_{i-1}<y<x_i\}$.
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 $\sup\{[a,b]\} \geq \sup\{(a,b)\}$.
 A: The definitions are not exactly equivalent. The $U^0$ definition can 'avoid' a finite number of points at which $f$ could be infinite, whereas if $f$ is infinite at a single point, then the $U$ integral will be infinite.
However, if $f$ is bounded, then the integrals are the same. So, I will assume $f$ is bounded on $[a,b]$.
First, it should be clear that $U^0(f;P) \leq U(f;P)$ (always, even if $f$ is unbounded).
Now let $P = (a=x_0,x_1,...,x_n=b)$ be a partition. Let $\epsilon>0$ such that $x_{k}+\epsilon < x_{k+1}-\epsilon$, and let $P_\epsilon = (x_0,x_0+\epsilon, x_1-\epsilon, x_1+\epsilon,...,x_n-\epsilon, x_n)$. Then
\begin{eqnarray}
U(f;P_\epsilon) &=& \sup_{x \in [x_0,x_0+\epsilon]} f(x) \epsilon + \sup_{x \in [x_0+\epsilon,x_1-\epsilon]} f(x) (x_1-x_0-2 \epsilon)+\cdots \\
&\leq& \sup_{x \in [x_0,x_0+\epsilon]} f(x) \epsilon + \sup_{x \in (x_0,x_1)} f(x) (x_1-x_0)+\cdots \\
& = & U^0(f;P)+ 2\epsilon(\sup_{x \in [x_0,x_0+\epsilon]} f(x) + \sup_{x \in [x_1-\epsilon,x_1+\epsilon]} f(x) \cdots) \\ 
& \leq & U^0(f;P)+ 2\epsilon (n+1) \sup_{x \in [x_0,x_1]} f(x)
\end{eqnarray}
So, in particular, we have for all $\epsilon>0$, for all partitions $P$, there exists a refinement $P'$ such that $U(f;P') \leq U^0(f;P)+\epsilon$.
 If $\cal P$ is the collection of partitions of $[a,b]$, it follows that $\inf_{P \in \cal{P}} U(f;P) =\inf_{P \in \cal{P}} U^0(f;P) $.
