Direct ways

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March 13th 2009, 07:27 AM
manjohn12

Direct ways

Are there any direct ways of proving that if $f$ is real-valued and Riemann integrable on $[a,b]$ , then $f$ is bounded?

Because the usual proof would be by contraposition:

Outline of "Usual Proof"

1. There exists a sequence $(y_n)$ in $[a,b]$ that converges to $y \in [a,b]$ , such that for every $n \in \mathbb{N}$ , $|f(y_n)| >n$ .

2. Let $P$ be a partition of $[a,b]$ . Then the set of Riemann sums of $f$ corresponding to $P$ is an unbounded set of real numbers.

3. Thus $f$ cannot be Riemann integrable on $[a.b]$ .
March 13th 2009, 08:02 AM
ThePerfectHacker

Quote:

Originally Posted by manjohn12

Are there any direct ways of proving that if $f$ is real-valued and Riemann integrable on $[a,b]$ , then $f$ is bounded?

This question seems strange to me. In the very definition of what it means for a function to be Riemann integrable it is assumed the function is bounded. Thus, there is no proof that the function is bounded, that is an assumption that we make about the function in the very beginning.
March 13th 2009, 08:04 AM
manjohn12

Quote:

Originally Posted by ThePerfectHacker

This question seems strange to me. In the very definition of what it means for a function to be Riemann integrable it is assumed the function is bounded. Thus, there is no proof that the function is bounded, that is an assumption that we make about the function in the very beginning.

The function is not assumed to be bounded in my definition.
March 13th 2009, 08:37 AM
manjohn12

Quote:

Originally Posted by ThePerfectHacker

This question seems strange to me. In the very definition of what it means for a function to be Riemann integrable it is assumed the function is bounded. Thus, there is no proof that the function is bounded, that is an assumption that we make about the function in the very beginning.

Here is the definition: Let $a,b \in \mathbb{R}$ with $a<b$ . Let $f$ be a real valued function whose domain consists of the interval $[a,b]$ . We say that $f$ is Riemann integrable on $[a,b]$ if there exists a real number $I$ such that for all $\epsilon >0$ , there exists $\delta >0$ such that $|\mathcal{R}(f,P)-I| < \epsilon$ whenever $\mathcal{R}(f,P)$ is a Riemann sum for $f$ corresponding to a partition of $[a,b]$ with mesh less than $\delta$ .
March 13th 2009, 09:19 AM
Plato

Quote:

Originally Posted by manjohn12

Here is the definition: Let $a,b \in \mathbb{R}$ with $a<b$ . Let $f$ be a real valued function whose domain consists of the interval $[a,b]$ . We say that $f$ is Riemann integrable on $[a,b]$ if there exists a real number $I$ such that for all $\epsilon >0$ , there exists $\delta >0$ such that $|\mathcal{R}(f,P)-I| < \epsilon$ whenever $\mathcal{R}(f,P)$ is a Riemann sum for $f$ corresponding to a partition of $[a,b]$ with mesh less than $\delta$ .

I will give you only an outline of a direct proof. A complete proof is messy with loads of subscripts. There is a partition of $[a,b]$ , $P = \left\{ {a = t_0 < t_1 < \cdots < t_n = b} \right\}$ such that $\left| {t_j - t_{j - 1} } \right| < \delta$ .
Suppose we have a refinement $Q \subseteq P$ then $\left| {\mathcal{R} (f;P) - \mathcal{R} (f;Q)} \right| < 1$ .
Now here comes the messy part.
By using right hand sums with respect to $P$ we change one number in the $Q$ sum $y \in \left[ {t_{j - 1} ,t_j } \right]$ we get $\begin{gathered} \left| {\left( {f(y) - f(t_j )} \right)\left( {t_j - t_{j - 1} } \right)} \right| \leqslant \left| {\left( {f(y) - f(t_j )} \right)} \right|\delta < 1 \hfill \\<br /> \left| {f(y)} \right| < \frac{1}<br /> {\delta } + \left| {f(t_j )} \right| \hfill \\ \end{gathered}$ .
Use the number $B = \sum\limits_{j = 1}^n {\frac{1}{\delta } + \left| {f(t_j )} \right|}$ for the bound.

I hope you can at least see how it might work.
March 13th 2009, 09:22 AM
ThePerfectHacker

Quote:

Originally Posted by manjohn12

Here is the definition: Let $a,b \in \mathbb{R}$ with $a<b$ . Let $f$ be a real valued function whose domain consists of the interval $[a,b]$ . We say that $f$ is Riemann integrable on $[a,b]$ if there exists a real number $I$ such that for all $\epsilon >0$ , there exists $\delta >0$ such that $|\mathcal{R}(f,P)-I| < \epsilon$ whenever $\mathcal{R}(f,P)$ is a Riemann sum for $f$ corresponding to a partition of $[a,b]$ with mesh less than $\delta$ .

Let $\epsilon=1$ and let $P=\{x_0,x_1,...,x_n\}$ . The function is unbounded on $[a,b]$ , therefore it must be unbounded on one of the subintervals: $[x_0,x_1],[x_1,x_2],...,[x_{n-1},x_n]$ . Say that $f$ is unbounded on $[x_{j-1},x_j]$ . Remember that $\mathcal{R}(f,P) = \sum_{k=1}^n f(t_k)\Delta_k$ where $t_k \in [x_{k-1},x_k]$ is some chosen point and $\Delta_k = x_k - x_{k-1}$ . Choose the points $t_0,...,t_{j-1},t_{j+1},...,t_n$ any way you want to. Then since $f$ is unbounded on $[x_{j-1},x_j]$ you can pick $t_j$ so that $\left| \sum_{k=1}^n f(t_k)\Delta_k\right| > M$ for any $M>0$ . This means if $|\mathcal{R}(f.P) - I | < 1\implies |\mathcal{R}(f,P) | < |I| + 1$ but we shown this is impossible.

EDIT: Indirect proof.
March 13th 2009, 09:37 AM
Plato

TPH. Note that manjohn12 asked for a direct proof.
He said that he knew several proofs by contradiction.
March 13th 2009, 06:03 PM
manjohn12

Quote:

Originally Posted by manjohn12

Are there any direct ways of proving that if $f$ is real-valued and Riemann integrable on $[a,b]$ , then $f$ is bounded?

Because the usual proof would be by contraposition:

Outline of "Usual Proof"

1. There exists a sequence $(y_n)$ in $[a,b]$ that converges to $y \in [a,b]$ , such that for every $n \in \mathbb{N}$ , $|f(y_n)| >n$ .

2. Let $P$ be a partition of $[a,b]$ . Then the set of Riemann sums of $f$ corresponding to $P$ is an unbounded set of real numbers.

3. Thus $f$ cannot be Riemann integrable on $[a.b]$ .

Off topic: For (1) All you need to show is that a a sequence exists that satisfies those properties? You use the compactness of $[a,b]$ to construct the sequence?