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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 $\sup\{[a,b]\} \geq \sup\{(a,b)\}$.

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1 Answer 1

up vote 2 down vote accepted

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) $.

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In line 2 of your inequalities are you missing an $\epsilon$? –  emka Nov 3 '12 at 9:29
    
@emka: I don't think so, the idea is to replace $\sup_{x \in [x_0+\epsilon,x_1-\epsilon]} f(x) (x_1-x_0-2 \epsilon)$ by $\sup_{x \in [x_0,x_1]} f(x) (x_1-x_0)$. Is this where you think the $\epsilon$ is missing? –  copper.hat Nov 3 '12 at 15:28
    
@cooper.hat Sorry it took me a while to get to this. I posted that only after a cursory reading. After a careful reading I understood everything. –  emka Nov 4 '12 at 9:05
    
No problem, glad to help! –  copper.hat Nov 4 '12 at 9:15
1  
There are two groups of terms on line 3, 'small' brackets around points of the partition $P$, and 'the rest'. $U^0(f;P)$ is larger than 'the rest' and the $2 \epsilon(\cdots)$ term upper bounds the sum of the 'small' brackets. Line 4 is just a coarse upper bound on the latter sum. Hopefully this clarifies... –  copper.hat Nov 5 '12 at 2:16

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