# Thread: Riemann Integrable Function

1. ## Riemann Integrable Function

Suppose f and g are Riemann Integrable Functions on[a,b] and f(x)=g(x) if x not equal to c.
show that : $
\int ^a_b f= \int ^a_b g
$

2. I googled away to glory and maybe this will help:
Darboux theory

EDIT: Can the experienced answerers shed light on this problem?

3. My advice would be to split the interval up into the following parts:
$[a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b]$.

The integrals on the last 2 parts are the same. For the region in the middle since $f,g$ are assumed Riemann integrable they are bounded. So say $|f|\le F, |g|\le G$. Thus you have that

$\int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|$

4. where the $c$ coming from?

5. Originally Posted by hkito
where the $c$ coming from?
Well c was given in the problem!
Do you understand the question?

6. Well I don't see any $c$ in the problem. It's just saying that for $c$ not equal to x. For me it's not clear enough.

7. It says that as long as x does not equal c you know that f(x)=g(x). However, you are not guaranteed that f(c)=g(c).

8. Okay,
so the problem is show that $\int f=\int g$ given that $f(x)=g(x)$ except at $x=c$

9. Originally Posted by putnam120
My advice would be to split the interval up into the following parts:
$[a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b]$.

The integrals on the last 2 parts are the same. For the region in the middle since $f,g$ are assumed Riemann integrable they are bounded. So say $|f|\le F, |g|\le G$. Thus you have that

$\int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|$

Since $f(x)=g(x)$ on $[a,c-\epsilon]\cup[c+\epsilon]$
you just need to show that $\int_{c-\epsilon}^{c+\epsilon}f=\int_{c-\epsilon}^{c+\epsilon}g$
For that you since you know that f and g are integrable, so $f-g$ is bounded , say by some $M$.

$\int_{c-\epsilon}^{c+\epsilon}|f(x)-g(x)|dx\leq M\int_{c-\epsilon}^{c+\epsilon}dx=2M\epsilon$
And you make $\epsilon$ go to zero

10. There's a really handy theorem in my book that says that "if a function has finitely many points of discontinuity, then it is Riemann-integrable". That's very nice for this problem, because if $h=f-g$, then $h=\left\{\begin{array}{lr}k&:x=c\\0&:x\neq c\end{array}\right\}$

Calculate the lower Riemann sum of $h$. Let $a=x_1 be a partition of $[a,b]$ and let $[x_{i-1},x_i]$ be the segment containing $c$. $\min h(x)$ on $[x_{i-1},x_i] = 0$, because it contains a point that is not $c$, so therefore the lower Riemann sum and the whole integral is zero.

11. Originally Posted by redsoxfan325
There's a really handy theorem in my book that says that "if a function has finitely many points of discontinuity, then it is Riemann-integrable".
If I remember correctly this can be extended to saying that if a function is discontinuous on a set of measure 0 then it is Riemann-integrable.

12. Originally Posted by putnam120
If I remember correctly this can be extended to saying that if a function is discontinuous on a set of measure 0 then it is Riemann-integrable.
(If this is true - it seems to be, but I don't have a proof lol) then it has to be with Lebesgue measure, isn't it ?

13. Yes sorry for the ambiguity.

14. We can prove something stronger. If $f$ is integrable and $g$ is a function that differs at countably many points distinct from $f$ such that they converge then $g$ is integrable with same value as $f$. The nice thing is that the proof is similar as in the finite case. Because when you have convergence it means that within a a small enough neighborhood of the limit all the points lie there. So you form a partition with thin enough rectangles around these values.