# Thread: Integral proof

1. ## Integral proof

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

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

Here some of my (incomplete) thoughts:

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

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

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

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

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

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

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

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

I am not sure how to introduce the limiting variable, $n$. 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 $\epsilon=\frac{1}{n}?$

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.

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

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

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