Proofing Change of Variable Formula for Integration

[sbashrawi]

Homework Statement

I need to proof the change of variable formula for integartion
integration of [g(t)]dt on [a+h, b+h] =integration of g(t+h)dt on [ a, b]

Homework Equations

The Attempt at a Solution

 

[╔(σ_σ)╝]

Can't you use a substitution?

 

[sbashrawi]

No, I don't think:

g( t) : [a+h, b+h]
g(t+h) : [a, b].

substitution will give : let x = t+h , then t = x-h which is defferent from the limits we have

 

[╔(σ_σ)╝]

u = t - h

du = dt

u_2 = (b+h) - h

u_1 = (a+h) - h

u is a dummy variable

The integral becomes

int g(u+h) on [a,b]

 

[sbashrawi]

I don't think this works , since you used two dumy ( u_ 1, u_2) variables to find the limits
and we are supposed to use just one

 

[micromass]

he didnt use dummy variables. u_1 is simply the llower bound of the domain of integration and u_2 is the upper bound...

 

[╔(σ_σ)╝]

I only used one dummy variable,u.

I have not idea what you mean by I used two.

I used u_2 and u_1 to state the limits of integration.

U_2 is simply the upper limit, u_1 respectively.

Edit
micromass to the rescue XD.

 

[sbashrawi]

I am sorry, You are right.

Thank you

 

[ColdCoffee]

I would have to think about it a little more, but my gut instinct is that just using the strait definition of the integral(I assume this is the Riemann integral right?) would be the most strait forward approach since you are just talking about a "linear" translation on the domain.

I don't think that this is the only way to do it but it is the one that my instinct tells me to use.

If I think of something I will let you know.

 

[sbashrawi]

In fact it is lebesgue integrable function and this is part of a problem. The problem asked to show this property for simple integrable function over [ a +h, b+h], then proceed to prove the general case.

 

[micromass]

Then there are 3 steps you need to do:

1) prove this for simple functions (i.e. sum of indicator functions)
2) by a limit argument, prove this for positive functions
3) prove the general case

 

[sbashrawi]

I know this but the problem is how to prove it.
I proved it in the follwoing way:
let f be simple function on [ a+h, b+h]
f(x) = sum( c_i X(E_i+h))
int f(t) over [a+h, b+h] = sum (c_i * m(E_i + h) = sum (c_i * m(E_i)) = sum (c_i * m( E_i -h))
= int f(t+h) over [a,b].

Am I right?

 

[ColdCoffee]

I know this but the problem is how to prove it.
I proved it in the follwoing way:
let f be simple function on [ a+h, b+h]
f(x) = sum( c_i X(E_i+h))
int f(t) over [a+h, b+h] = sum (c_i * m(E_i + h) = sum (c_i * m(E_i)) = sum (c_i * m( E_i -h))
= int f(t+h) over [a,b].

Am I right?

I like the principle of what you are doing (IE, using the fact that the measure of a subset of reals is invariant under under "linear" shifts). I would check your shift again again(it looks like you shifted by h twice). Also, make sure that it is clear in the context of your class that those integrals are in fact correct. IE you have completely changed your simple function- how do you know that the integrals are really the same? (Likely this is clear in your context- I mention it because my Measure Theory prof was super attentive to detail- not justifying that would have cost me at least 20% of the points for the exercise).