# Starting point for Riemann sum, Integral proof?

• Feb 9th 2010, 08:08 AM
paupsers
Starting point for Riemann sum, Integral proof?
Suppose f is integrable on the interval [a,b] and Q is a partition of [a,b]. Suppose $\epsilon >0$. Prove that there exists $\delta >0$ such that if P is any partition of [a,b] with $|P|<\delta$ and S(P,f) is a Riemann sum then

$|S(P,f)-\int_a^bf|<\epsilon$

I'm mostly just looking for a starting off point, or a direction to go. Should I try proving that $S(P,f)=\int_a^bf$ for $\delta$ small enough? How should I define $\delta$?
• Feb 9th 2010, 07:39 PM
paupsers
• Feb 9th 2010, 08:05 PM
Drexel28
Quote:

Originally Posted by paupsers
Suppose f is bounded on the interval [a,b] and Q is a partition of [a,b]. Suppose $\epsilon >0$. Prove that there exists $\delta >0$ such that if P is any partition of [a,b] with $|P|<\delta$ and S(P,f) is a Riemann sum then

$|S(P,f)-\int_a^bf|<\epsilon$

I'm mostly just looking for a starting off point, or a direction to go. Should I try proving that $S(P,f)=\int_a^bf$ for $\delta$ small enough? How should I define $\delta$?

what does $|P|$ mean? Is it the mesh?
• Feb 9th 2010, 09:02 PM
paupsers
Yes.
• Feb 10th 2010, 03:16 PM
paupsers
Still trying to figure this one out. Anyone have any ideas?
• Feb 10th 2010, 03:30 PM
Plato
Quote:

Originally Posted by paupsers
Suppose f is bounded on the interval [a,b] and Q is a partition of [a,b]. Suppose $\epsilon >0$. Prove that there exists $\delta >0$ such that if P is any partition of [a,b] with $|P|<\delta$ and S(P,f) is a Riemann sum then

$|S(P,f)-\int_a^bf|<\epsilon$

I'm mostly just looking for a starting off point, or a direction to go. Should I try proving that $S(P,f)=\int_a^bf$ for $\delta$ small enough? How should I define $\delta$?

As stated the statement is false. That is why on one has answered.
It seems to say that if a function is bounded on a closed interval it is Riemann integrable on the interval.
That is clearly not true.

On the other hand, if you mean that the function is bounded and Riemann integrable then the conclusion is the very definition of Riemann integrable. So there is nothing to prove, unless you are using a different definition. Are you?
If so, what is it?

I suggest that you review the posting and make adjustments.
• Feb 10th 2010, 05:13 PM
paupsers
Yes, you're right, part of the hypothesis should be that the function is Riemann integrable.

Does this proof work?

Proof: We know $L(P,f)\leq S(P,f)\leq U(P,f)$
We also know that $L(P,f)\leq \int_a^bf\leq U(P,f)$
This implies that $L(P,f)\leq |S(P,f)-\int_a^bf|\leq U(P,f)$
Since f is integrable on [a,b], we know there exists a partition P such that $U(P,f)-L(P,f)<\epsilon$.
This implies that $|S(P,f)-\int_a^bf|<\epsilon$. QED.

• Feb 10th 2010, 05:21 PM
Drexel28
Quote:

Originally Posted by paupsers
Yes, you're right, part of the hypothesis should be that the function is Riemann integrable.

Does this proof work?

Proof: We know $L(P,f)\leq S(P,f)\leq U(P,f)$
We also know that $L(P,f)\leq \int_a^bf\leq U(P,f)$
This implies that $L(P,f)\leq |S(P,f)-\int_a^bf|\leq U(P,f)$
Since f is integrable on [a,b], we know there exists a partition P such that $U(P,f)-L(P,f)<\epsilon$.
This implies that $|S(P,f)-\int_a^bf|<\epsilon$. QED.

I think the problem for me is I don't know what the hell $S(P,f)$ is. But, assuming that $L(P,f)\leqslant S(P,f)\leqslant U(P,f)$....since $f$ is integrable $L(P,f)\leqslant \int_a^b\text{ }f\text{ }dx\leqslant U(P,f)$. So, $\int_a^b\text{ }f\text{ }dx-S(P,f)\leqslant U(P,f)-L(P,f)$ and equally valid we see that $S(P,f)-\int_a^b\text{ }f\text{ }dx\leqslant U(P,f)-L(P,f)$. Combining these we get $\left|\int_a^b\text{ }f\text{ }dx-S(P,f)\right|\leqslant U(P,f)-L(P,f)$ for every partition $P$. Thus, since $f$ is Riemann integrable we see that for every $\varepsilon>0$ there exists some $P\in\wp$ such that $\varepsilon>U(P,f)-L(P,f)<\left|\int_a^b\text{ }f\text{ }dx-S(P,f)\right|$ and teh conclusion follows.
• Feb 11th 2010, 07:10 AM
Plato
Quote:

Originally Posted by paupsers
We know $L(P,f)\leq S(P,f)\leq U(P,f)$
We also know that $L(P,f)\leq \int_a^bf\leq U(P,f)$
This implies that $\color{red}L(P,f)\leq |S(P,f)-\int_a^bf|\leq U(P,f)$
This implies that $|S(P,f)-\int_a^bf|<\epsilon$. QED.

The part is red does not follow at once.
But we can say $S(P,f)\leq U(P,f)$ and $-\int_a^bf\leq -L(P,f)$.
Then $0\le S(P,f)-\int_a^bf\leq U(P,f)-L(P,f)$.
Now finish.