Riemann Integrable Function

**math43** · May 3rd 2009, 10:53 PM

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

**fardeen_gen** · May 8th 2009, 09:59 PM

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

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

**putnam120** · May 9th 2009, 08:55 AM

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

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

**hkito** · May 9th 2009, 04:06 PM

where the $c$ coming from?

**Plato** · May 9th 2009, 04:40 PM

Originally Posted by hkito

where the $c$ coming from?

Well c was given in the problem!
Do you understand the question?

**hkito** · May 9th 2009, 04:46 PM

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.

**putnam120** · May 9th 2009, 04:48 PM

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).

**hkito** · May 9th 2009, 04:56 PM

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

**hkito** · May 9th 2009, 05:11 PM

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

**redsoxfan325** · May 9th 2009, 10:57 PM

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 $a=x_1<x_2<...<x_n=b$ 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.

**putnam120** · May 10th 2009, 12:19 PM

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.

**Moo** · May 10th 2009, 12:48 PM

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 ?

**putnam120** · May 10th 2009, 01:06 PM

Yes sorry for the ambiguity.

**ThePerfectHacker** · May 10th 2009, 05:47 PM

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.

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