Solving Integral Proof: Let a<b<c, f:[a,c]->R

[e(ho0n3]

I'm having a hard time understanding the proof of the following: Let a < b < c and let f: [a,c] -> R be Riemann integrable on [a,b], [b,c] and [a,c]. Then

$$\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx$$.

Proof. Let C and C' be the characteristic functions of [a,b] and [b,c] respectively, defined on [a,c]. Then f = C f + C' f and the addition formula above follows from the linearity of the integral.

This is such a facile proof. Sigh. I'm trying to fill in the missing details: I know that since f = C f + C' f, then

$$\int_a^c f(x) \, dx = \int_a^c C(x) f(x) \, dx + \int_a^c C'(x) f(x) \, dx$$

where I've used the fact that characteristic functions are Riemann integrable and products of Riemann integrable functions are Riemann integrable. Now how would I show, without much fuss, that

$$\int_a^c C(x) f(x) \, dx = \int_a^b f(x) \, dx$$

for example?

 

[mathman]

I'm having a hard time understanding the proof of the following: Let a < b < c and let f: [a,c] -> R be Riemann integrable on [a,b], [b,c] and [a,c]. Then

$$\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx$$.

Proof. Let C and C' be the characteristic functions of [a,b] and [b,c] respectively, defined on [a,c]. Then f = C f + C' f and the addition formula above follows from the linearity of the integral.

This is such a facile proof. Sigh. I'm trying to fill in the missing details: I know that since f = C f + C' f, then

$$\int_a^c f(x) \, dx = \int_a^c C(x) f(x) \, dx + \int_a^c C'(x) f(x) \, dx$$

where I've used the fact that characteristic functions are Riemann integrable and products of Riemann integrable functions are Riemann integrable. Now how would I show, without much fuss, that

$$\int_a^c C(x) f(x) \, dx = \int_a^b f(x) \, dx$$

for example?

Just use the fact that C(x)=0 in the interval [b,c], so that C(x)f(x)=0 in this interval, and its integral=0.

 

[e(ho0n3]

Just use the fact that C(x)=0 in the interval [b,c], so that C(x)f(x)=0 in this interval, and its integral=0.

Where do I use that fact exactly?

 

[Vid]

Adding what mathman said to the end of the proof would be sufficient, but a rigorous proof could be done using upper and lower sums.

 

[e(ho0n3]

Yeah. I decided to ditch that proof. I found a more elementary but longer proof using the definition of Riemann integration, which I found satisfactory.

 

[morphism]

Why don't you find the proof in the OP satisfactory?

 

[e(ho0n3]

As I wrote, I don't know why

$$\int_a^c C(x) f(x) \, dx = \int_a^b f(x) \, dx$$

is true.

 

[morphism]

But you were given two justifications as to why that is true. The Riemann sums of C(x)f(x) will always be zero on [b,c].

 

[e(ho0n3]

Those two justifications did nothing for me. I was hoping someone would give a simple argument that would not use Riemann sums. Oh well. Thanks anyways.

 

[mathman]

Where do I use that fact exactly?

The integral from a to c is the sum of the integral from a to b and the integral from b to c. From a to b, C(x)=1, so the integral of Cf = integral of f. From b to c, C(x)=0, so Cf=0 and the integral = 0.

 

[e(ho0n3]

The integral from a to c is the sum of the integral from a to b and the integral from b to c. From a to b, C(x)=1, so the integral of Cf = integral of f. From b to c, C(x)=0, so Cf=0 and the integral = 0.

That's a circular argument: You're using the very fact I want to prove.

 

[sutupidmath]

I don't know whether you will find the following proof useful, if you haven't already seen it before, but here it is an elementary proof of

$$\int_a^c f(x) \, dx = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx$$

Proof:

Let a<c<b. Since the reiman sums do not depend on the way we may partition the interval [a,b] we can partition this interval first into two subintervals, let them

[a,c] and [c,b] where c is the same point we are using in the integral

Then we may do the following partitition to both intervals

$$a=x_0<x_1<x_2<...<x_k=c$$ and $$c=x_k<x_{k+1}<...<x_n=b$$

So, we can form the following integral sums for both intervals:


$$\sum_{i=1}^kf(\delta_i)\triangle x_i$$

$$\sum_{i=k}^nf(\delta_i)\triangle x_i$$

where $$\delta_i \in [x_{i-1},x_i]$$ and

$$\triangle x_i=x_i-x_{i-1}$$


Now, since we are dealing here with finite sums, then we have the following relation


$$\sum_{i=1}^nf(\delta_i)\triangle x_i=\sum_{i=1}^kf(\delta_i)\triangle x_i + \sum_{i=k}^nf(\delta_i)\triangle x_i$$

Now, if $$max \triangle x_i->0$$ then taking the limit on both sides,we get our desired result.

A simmilar argument follows in two other cases when c is not between a and b.

 

[mathman]

That's a circular argument: You're using the very fact I want to prove.

It is NOT circular. I don't see what is needed to prove, unless you need a proof that multiplication by 1 leaves the function unchange and multiplication by 0 results in 0.

 

[sutupidmath]

It is NOT circular. I don't see what is needed to prove, unless you need a proof that multiplication by 1 leaves the function unchange and multiplication by 0 results in 0.

I appologize to just throw stuff here, but i think that e(ho0n3's real question is what is a characteristic function. I think he is dealing with this problem, and hence it is not clear to him why C(x)=1 in [a,b] and C(x)=0 on [b,c]

I think that a characteristic function of an interval say

$$[x_1,x_2]$$

is as follows


$$C(x)=\left\{\begin{array}{cc}1,&\mbox{ if } x\in[x_1,x2]\\0, & \mbox{ if } x_2<x<x_1\end{array}\right.$$

So, this automatically would mean that if C(x) is your characteristic function on the interval [a,b] then C(x)=1 when x is in [a,b] and C(x)=0 when x is in [b,c] the same for C'(x)=1 if x is in [b,c] and C'(x)=0 when x is in [a,b]

P.s. I once more appologize for i am only a freshman, and my advice shall probably turn out to be helpless and inappropriate.

 

[sutupidmath]

It is NOT circular. I don't see what is needed to prove, unless you need a proof that multiplication by 1 leaves the function unchange and multiplication by 0 results in 0.

Well, these two would also follow from the axioms of a field, right?

Unless one wanted not to rely on algebra at all, but rather on dedekind cuts etc.