# Arithmetic of of integrable functions

1. Jun 21, 2005

### steven187

hello all

i have been trying to prove a property of integrable functions, i had a go at it dont know if it is correct, but im wondering if there could possibly be a shorter simpler way of proving it alright here we go

$$\int_{a}^{b} f(x)+g(x) dx= \int_{a}^{b} f(x)dx +\int_{a}^{b} g(x) dx$$

My proof

for any partition P of [a,b]

$$U(f+g,P)=\sum_{i=1}^{n}M_i(f+g,P)(x_i-x_{i-1})$$

$$\le \sum_{i=1}^{n}M_i(f,P)(x_i-x_{i-1}) +\sum_{i=1}^{n}M_i(f,P)(x_i-x_{i-1})$$

$$= U(f,P)+U(f,g)$$ similarly

$$L(f+g,P)\ge L(f,P)+L(f,g)$$

there is also partitions $$P_{1}$$ & $$P_{2}$$ of [a,b] such that

$$U(f,P_{1}) <\int_{a}^{b}f+\frac{\epsilon}{2}$$

$$U(g,P_{2}) <\int_{a}^{b}g+\frac{\epsilon}{2}$$

we let $$P=P_{1}UP_{2}$$

$$\int_{a}^{b^U}(f+g)\le U(f+g,P)\le U(f,P)+U(g,P) \le U(f,P_{1})+U(g,P_{2})$$

$$< \int_{a}^{b}f+\frac{\epsilon}{2}+\int_{a}^{b}g+\frac{\epsilon}{2}$$

$$= \int_{a}^{b}f+\int_{a}^{b}g+\epsilon$$
similarly
$$\int_{a_{L}}^{b}(f+g) > \int_{a}^{b}f+\int_{a}^{b}g -\epsilon$$
since all functions are riemann integrable then
$$\int_{a}^{b^U}(f+g) =\int_{a_{L}}^{b}(f+g)= \int_{a}^{b}(f+g)$$
and so it follows that
$$\mid\int_{a}^{b}(f+g)-\int_{a}^{b}f-\int_{a}^{b}g\mid<\epsilon$$ $$\forall\epsilon>0$$

Last edited: Jun 21, 2005
2. Jun 21, 2005

### matt grime

most iof thsat seems unnecessary

if U(f,P)+U(g),P) +> U(f+g,P) => L(f+g,P) => L(f,P)+L(g,P) then the result follows since f and g are integrable

3. Jun 21, 2005

### steven187

hello there

well i have looked at your approach but i dont understand i would appreciate it if you can elaborate,

thank you

4. Jun 21, 2005

### steven187

hello all

also is there any special way of doing such proofs with riemann integrability or is there something I would need to know that is in common with most proofs involving the riemann integral please help its just im finding it difficult to approach such proofs

thank you

5. Jun 21, 2005

### matt grime

the limits you want to find anre sandwihced between two convergent "sequences". f and g are integrable so the upper and lower sums exist and converge; apply the sandwich principle.

6. Jun 21, 2005

### steven187

hello there

hmm..well I dont get what you mean by limits and convergent sequences, im getting really confused in how to relates those terms to proving this problem, I want to work on another approach of doing this problem but I dont know where else to start, also is there any special way of doing such proofs with riemann integrability or is there something I would need to know that is in common with most proofs involving the riemann integral please help its just im finding it difficult to approach such proofs, please help

thank you

7. Jun 21, 2005

### matt grime

there are exactly tow =thnigs to know

1. the definition

2 the criterion of integrability

i have broken my right hand and find typing a pain at the mo'. hopefuly someone else can explain., but you do undrestand that the integral exists if the lower and upper sums converge as the parto=itions longst subpatrition tends to zero and that it suffices to prove this only for a sequence og partitions indexed by N. didn't we prove that for yuou already?

8. Jun 21, 2005

### fourier jr

start wity matt's line. then write down riemann's criterion for integrability, using epsilon/2 for each of f, g. conclude that f + g is riemann integrable by riemann's criterion. then use matt's line & properties of partitions to find out what f + g is.

9. Jun 21, 2005

### Hurkyl

Staff Emeritus
(Assuming the errors in your original post are typos)

You've proven that for any partition P:

U(f + g, P) <= U(f, P) + U(g, P)
and
L(f, P) + L(g, P) <= L(f + g, P)

right?

What do you know about, say, L(f, P) and U(f, P) as P &rarr; &infin;?

10. Jun 21, 2005

### steven187

hello all

yep thats what i have proved
Well Matt I hope the hand gets better. now from my understanding if the lower and upper sums converge that is as the number of partitions approaches infinity and the subintervals get smaller and smaller, and so if the upper and lower sums converge to the same value then it is riemann integrable.

the only information i can gather is what i have written originally in my proof above and
U(f,P)-L(f,P) <e/2
U(g,P)-L(g,P) <e/2
U(f+g,P)-L(f+g,P) <= U(f,P)-L(f,P)+U(g,P)-L(g,P)<e/2+e/2=e
now from here i dont know what else to do except to go along the same path as the original way i proved it,

11. Jun 21, 2005

### Hurkyl

Staff Emeritus
One thing you can do is take the limit of the inequalities

U(f + g, P) <= U(f, P) + U(g, P)
L(f, P) + L(g, P) <= L(f + g, P)

as |P| → ∞

12. Jun 21, 2005

### steven187

dont you mean |P| → 0
isnt the maximum difference between any two consecutive points of the partition is called the norm or mesh of the partition and is denoted by |P|

$$\int_{a}^{b}f+\int_{a}^{b}g\le\int_{a_{L}}^{b}(f+g) = \int_{a}^{b}(f+g) =\int_{a}^{b^U}(f+g) \le\int_{a}^{b}f+\int_{a}^{b}g$$

would this be correct, is there anything else that i need to add?

13. Jun 21, 2005

### steven187

hello all

by the way, would i be correct to say that there are only 3 ways to prove one expression is equal to the other and they would be
sandwich theorem a<=b<=a hence a=b
LHS=RHS
|a-b|<e for all e>0 then hence a=b
would there be any other ways? and would anybody know of any links that would explain any other forms of proving inequality?

thanxs