Suppose f and g are Riemann Integrable Functions on[a,b] and f(x)=g(x) if x not equal to c.

show that : $\displaystyle

\int ^a_b f= \int ^a_b g

$

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- May 3rd 2009, 10:53 PMmath43Riemann 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 : $\displaystyle

\int ^a_b f= \int ^a_b g

$ - May 8th 2009, 09:59 PMfardeen_gen
I googled away to glory and maybe this will help:

**Darboux theory**

EDIT: Can the experienced answerers shed light on this problem? - May 9th 2009, 08:55 AMputnam120
My advice would be to split the interval up into the following parts:

$\displaystyle [a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b]$.

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

$\displaystyle \int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|$ - May 9th 2009, 04:06 PMhkito
where the $c$ coming from?

- May 9th 2009, 04:40 PMPlato
- May 9th 2009, 04:46 PMhkito
Well I don't see any $\displaystyle c $ in the problem. It's just saying that for $\displaystyle c$ not equal to x. For me it's not clear enough.

- May 9th 2009, 04:48 PMputnam120
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).

- May 9th 2009, 04:56 PMhkito
Okay,

so the problem is show that $\displaystyle \int f=\int g $ given that $\displaystyle f(x)=g(x)$ except at $\displaystyle x=c$ - May 9th 2009, 05:11 PMhkito

Since $\displaystyle f(x)=g(x)$ on $\displaystyle [a,c-\epsilon]\cup[c+\epsilon]$

you just need to show that $\displaystyle \int_{c-\epsilon}^{c+\epsilon}f=\int_{c-\epsilon}^{c+\epsilon}g$

For that you since you know that f and g are integrable, so $\displaystyle f-g$ is bounded , say by some$\displaystyle M$.

$\displaystyle \int_{c-\epsilon}^{c+\epsilon}|f(x)-g(x)|dx\leq M\int_{c-\epsilon}^{c+\epsilon}dx=2M\epsilon$

And you make $\displaystyle \epsilon$ go to zero - May 9th 2009, 10:57 PMredsoxfan325
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 $\displaystyle h=f-g$, then $\displaystyle h=\left\{\begin{array}{lr}k&:x=c\\0&:x\neq c\end{array}\right\}$

Calculate the lower Riemann sum of $\displaystyle h$. Let $\displaystyle a=x_1<x_2<...<x_n=b$ be a partition of $\displaystyle [a,b]$ and let $\displaystyle [x_{i-1},x_i]$ be the segment containing $\displaystyle c$. $\displaystyle \min h(x)$ on $\displaystyle [x_{i-1},x_i] = 0$, because it contains a point that is not $\displaystyle c$, so therefore the lower Riemann sum and the whole integral is zero. - May 10th 2009, 12:19 PMputnam120
- May 10th 2009, 12:48 PMMoo
- May 10th 2009, 01:06 PMputnam120
Yes sorry for the ambiguity.

- May 10th 2009, 05:47 PMThePerfectHacker
We can prove something stronger. If $\displaystyle f$ is integrable and $\displaystyle g$ is a function that differs at countably many points distinct from $\displaystyle f$ such that they converge then $\displaystyle g$ is integrable with same value as $\displaystyle f$. 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.