# Thread: Integral at Finite Points

1. ## Integral at Finite Points

For homework we had the following problem which I actually found difficult and fun. (Ross, Section 32, Problem 7).

"Let f be an integrable function on [a,b] and g be a function which agrees with f except for finitely many points, show that g is also integrable and furthermore:
INT(a,b;f) = INT(a,b;g)"

My book (in the back) gives a sketch of the solution. Which is a nightmare.

I came up with two solutions, one of which the professor did in class. But I did not like it. I will explain why. He used the following theorem:

Theorem: If f and g are integrable on [a,b] then f+g is too and furthermore:
INT(a,b;f)+INT(a,b;g) = INT(a,b;f+g)

The reason why I did not like it is because it uses a theorem which was not proved, i.e. from section 33.

See if you propose another proof without that. I found it fun. Note: We use the Darboux definition of the integral.

2. Originally Posted by ThePerfectHacker
For homework we had the following problem which I actually found difficult and fun. (Ross, Section 32, Problem 7).

"Let f be an integrable function on [a,b] and g be a function which agrees with f except for finitely many points, show that g is also integrable and furthermore:
INT(a,b;f) = INT(a,b;g)"

My book (in the back) gives a sketch of the solution. Which is a nightmare.

I came up with two solutions, one of which the professor did in class. But I did not like it. I will explain why. He used the following theorem:

Theorem: If f and g are integrable on [a,b] then f+g is too and furthermore:
INT(a,b;f)+INT(a,b;g) = INT(a,b;f+g)

The reason why I did not like it is because it uses a theorem which was not proved, i.e. from section 33.

See if you propose another proof without that. I found it fun. Note: We use the Darboux definition of the integral.
Why is this not trivial, as the points are in finitely many of the subintervals
of each partition and the contributions of the subintervals containing these
exceptional points goes to zero as the partition is refined (since the functions
are both integrable) the integrals must be equal.

or am I missing some fine point here?

RonL

3. Originally Posted by CaptainBlank
or am I missing some fine point here?
Yes. This problem is on the section on simply the definitions of the integral with along a condition for integrability. You are using the subdivision theorem which is not in the section.

4. I agree with RonL, this should be trivial.
Because there is a finite set in [a,b], we can find a finite collection of disjoint open intervals which covers that set. Let O be the union of the open intervals. We can make the total length, the measure, of O as small as we need. Moreover, g has a maximum on O. The set [a,b]\O is a collection of closed intervals such that on each f=g is integrable. Given any partition of [a,b], we can find a refinement relative to set O such that the difference in the f-sum and the g-sum is as small as needed.

5. Originally Posted by ThePerfectHacker
Yes. This problem is on the section on simply the definitions of the integral with along a condition for integrability. You are using the subdivision theorem which is not in the section.
You will have to explain in a bit more detail, I think we need only the definition
of the Darboux integral, or maybe you are working with an equivalent definition?

RonL

6. Originally Posted by CaptainBlank
You will have to explain in a bit more detail, I think I need only the definition
of the Darboux integral.
My analysis professor asked different students to do summaries of different sections. I did the one on integration (actually I traded with a classmate).

Now I would tell you the site, but you need to registered as a college student to do that.

So I will do that for you.

Heir.

7. Originally Posted by ThePerfectHacker
My analysis professor asked different students to do summaries of different sections. I did the one on integration (actually I traded with a classmate).

Now I would tell you the site, but you need to registered as a college student to do that.

So I will do that for you.