# Thread: Integral proof

1. ## Integral proof

Q: Prove that a bounded function $\displaystyle f$ is integrable on $\displaystyle [a,b]$ if and only if there exists a sequence of partitions $\displaystyle (P_{n})_{n=1}^{\infty}$ satisfying

$\displaystyle lim_{n\rightarrow\infty}[U(f,P_{n})-L(f,P_{n})]=0$.

Here some of my (incomplete) thoughts:

$\displaystyle (\Leftarrow)$
Let $\displaystyle \epsilon>0$. Since there exists a sequence of partitions $\displaystyle (P_{n})_{n=1}^{\infty}$ satisfying $\displaystyle lim_{n\rightarrow\infty}[U(f,P_{n})-L(f,P_{n})]=0$, there exists an $\displaystyle N$ such that, for any $\displaystyle n\geq\\N$ $\displaystyle U(f,P_{n})-L(f,P_{n})<\epsilon$. Thus, $\displaystyle f$ integrable on $\displaystyle [a,b]$.

$\displaystyle (\Rightarrow)$
For this direction I am not sure how to show there exists a sequence of partitions. Since we are assuming $\displaystyle f$ is integrable, I know that there is a partion $\displaystyle P_{\epsilon}$ of $\displaystyle [a,b]$ such that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$ for any positive $\displaystyle \epsilon$.

2. Using a the definition of integral if $\displaystyle f$ on $\displaystyle [a,b]$ then there is a number $\displaystyle I$ such that for any positive $\displaystyle \epsilon>0$ there is a partition $\displaystyle P$ of $\displaystyle [a,b]$ then $\displaystyle \left|\mathcal{S} (P;f)-I\right|<\epsilon$.

Now here is the trick. We can do it twice with $\displaystyle \frac{\epsilon}{2}$
$\displaystyle \left|\mathcal{U} (P;f)-I\right|<\frac{\epsilon}{2}$ and $\displaystyle \left|\mathcal{L} (P;f)-I\right|<\frac{\epsilon}{2}$.
Now complete it.

3. Define $\displaystyle U(f)=inf\{U(f,P):P\in\\Q\}$ and $\displaystyle L(f)=sup\{L(f,P):P\in\\Q\}$, where $\displaystyle Q$ is the set of all possible partitions.

We are given $\displaystyle f$ is integrable on $\displaystyle [a,b]$, so there exists a partition $\displaystyle P_{\epsilon}$ of $\displaystyle [a,b]$ such that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$. Since $\displaystyle f$ is integrable $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})=(U(f,P_{\epsilon})-U(f))+(L(f)-L(f,P_{\epsilon})$.

Let $\displaystyle I=U(f)=L(f)$. Then, $\displaystyle |U(f,P_{\epsilon})-I|<\frac{\epsilon}{2}$ and $\displaystyle |L(f,P_{\epsilon})-I|<\frac{\epsilon}{2}$.

So, we have a sequence of paritions $\displaystyle (P_{n})_{n=1}^{\infty}$ such that, there exists an $\displaystyle N>0$ for which any $\displaystyle n\geq\\N$ implies the upper and lower sums converge to the same value.

I am not sure how to introduce the limiting variable, $\displaystyle n$. I understand what is going on, but I am having trouble formalizing the idea of a sequence of partitions.

4. You cannot do the above for $\displaystyle \epsilon=\frac{1}{n}?$

5. Sorry to bring up an old thread, but I have been away for a while and have not yeat figured out the proof.

I am unsure how to prove the existance of the sequence of partitions.

$\displaystyle f$ is integrable on $\displaystyle [a,b]$, so for any $\displaystyle \epsilon>0$ we have that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$. Since $\displaystyle \epsilon>0$ is arbitrary it is certianly true that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})$ can be made less than $\displaystyle \epsilon=\frac{1}{n}$ for each $\displaystyle n\in\mathbb{N}$. Therefore, there exists a sequenct of paritions $\displaystyle P_{n}=\frac{b-a}{n}$ such that $\displaystyle n>N$, where $\displaystyle N\in\mathbb{N}$, implies $\displaystyle (P_{n})\rightarrow\\P_{\epsilon}$.

Notice,
$\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})=U(f,P_{\epsilon})-U(f)+L(f)-L(f,P_{\epsilon})$ since $\displaystyle f$ is integrable. Thus, $\displaystyle U(f,P_{\epsilon})-U(f)$ can be made less than half the prescribled $\displaystyle \epsilon$ as can $\displaystyle L(f)-L(f,P_{\epsilon})$. Since $\displaystyle U(f)=I=L(f)$, we have that the lower and upper sums converge to the same limit. Thus,

$\displaystyle U(f,P_{n})-L(f,P_{n})=0$ as $\displaystyle n\rightarrow\infty$.