Why do we take this partition?

[evinda]

Hey! :)
Let $f:[a,b] \to \mathbb{R}$ bounded.We suppose that f is continuous at each point of $[a,b]$,except from $c$.Prove that $f$ is integrable.

We suppose that $c=a$.
$f$ is bounded,so $\exists M>0$ such that $|f(x)|\leq M \forall x$

Let $\epsilon'>0$.We pick now a $x_0 \in (a,b)$ such that $x_0-a< \epsilon'$.

Let $\epsilon''>0$

As $f$ is continuous at $[x_0,b]$ ,it is integrable.

So there is a partition $P$ of $[x_0,b]$ such that: $U(f,P)-L(f,P)<\epsilon''$

Now we consider the partition $P'=\{a\} \cup P$ of $[a,b]$.But,why do we take this partition??Aren't there any other points between $a$ and $x_0$ ??

 

[Opalg]

The idea is to choose $x_0$ so close to $a$ that you do not need any other partition points between them.

You know that $f$ always has to lie between $-M$ and $+M$, so the difference between the upper and lower sums on the interval $[a,x_0]$ is at most $2M(x_0-a).$ Choose $x_0$ so close to $a$ that that difference is less than $\epsilon'$. Then, exactly as in your proof, choose a partition $P$ for the reminder of the interval so that $U(f,P)-L(f,P)<\epsilon''$. Define $P'=\{a\} \cup P$, then $U(f,P')-L(f,P')<\epsilon' + \epsilon''$. Since you can make that as small as you like, you have proved that $f$ is integrable on $[a,b].$

 

[evinda]

The idea is to choose $x_0$ so close to $a$ that you do not need any other partition points between them.

You know that $f$ always has to lie between $-M$ and $+M$, so the difference between the upper and lower sums on the interval $[a,x_0]$ is at most $2M(x_0-a).$ Choose $x_0$ so close to $a$ that that difference is less than $\epsilon'$. Then, exactly as in your proof, choose a partition $P$ for the reminder of the interval so that $U(f,P)-L(f,P)<\epsilon''$. Define $P'=\{a\} \cup P$, then $U(f,P')-L(f,P')<\epsilon' + \epsilon''$. Since you can make that as small as you like, you have proved that $f$ is integrable on $[a,b].$

I understand..Thank you very much! :)