How to Prove the Measure Property for a Nonnegative Measurable Function?

[Kindayr]

Homework Statement

Let (X,\mathcal{B},\mu) be a measure space and g be a nonnegative measurable function on X. Set \nu (E)=\int_{E}g\,d\mu. Prove that <br /> \nu is a measure and \int f\, d \nu =\int fg\,d\mu for all nonnegative measurable functions f on X.

The Attempt at a Solution

I'm basically at a total loss on how to start this. I'll keep working on it tonight, and will add to it as I go.

 

[morphism]

Have you tried applying the definition of "measure" to \nu? You'll have to use some basic results, like the monotone convergence theorem, to get stuff to work out.

 

[Kindayr]

I've shown for non-negative simple functions that \int \phi\,d\nu=\int \phi g \,d\mu. Now I wish to show it in general for non-negative measurable functions. So I say let f be a non-negative measurable function on X. Fix \phi as a simple function such that 0\le\phi \le f. Hence we have \int \phi\,d\nu=\int \phi g \,d\mu, and thus \int f\, d\nu=\sup_{\phi} \int \phi \, d\,\nu=\sup_{\phi} \int g\phi\, d\,\mu \overset{?}{=} \int gf\, d\mu.

Am I allowed to make that last equality?

 

[Kindayr]

Should I do some inequalities for simple functions above and below? I feel like that last equality should be an inequality. Hrmf.

 

[Ray Vickson]

Homework Statement

Let (X,\mathcal{B},\mu) be a measure space and g be a nonnegative measurable function on X. Set \nu (E)=\int_{E}g\,d\mu. Prove that <br /> \nu is a measure and \int f\, d \nu =\int fg\,d\mu for all nonnegative measurable functions f on X.

The Attempt at a Solution

I'm basically at a total loss on how to start this. I'll keep working on it tonight, and will add to it as I go.

Try it first for step functions f, of the form
f = \sum_{i=1}^n c_i \chi(I_i), where I_1, I_2,..., I_n is a measurable partition of R and \chi(A) is the characteristic function of a set A \subset R: \: \chi(A)(x) = 0 \text{ if } x \not\in A, \text{ and } \chi(A)(x) = 1 \text{ if } x \in A.

RGV

 

[Kindayr]

Try it first for step functions f, of the form
f = \sum_{i=1}^n c_i \chi(I_i), where I_1, I_2,..., I_n is a measurable partition of R and \chi(A) is the characteristic function of a set A \subset R: \: \chi(A)(x) = 0 \text{ if } x \not\in A, \text{ and } \chi(A)(x) = 1 \text{ if } x \in A.

RGV

Hey, thanks for the reply!

I've already done this for simple functions, I'm just stuck on how to show that any non-negative measurable function satisfies the equality. Should i take approximations from above and below? or should what I gave in my first reply be sufficient?

 

[Ray Vickson]

Hey, thanks for the reply!

I've already done this for simple functions, I'm just stuck on how to show that any non-negative measurable function satisfies the equality. Should i take approximations from above and below? or should what I gave in my first reply be sufficient?

I don't know how your textbook defines the integral for general functions, but many treatments define it as the limit of integrals of step functions.

RGV

 

[Kindayr]

I don't know how your textbook defines the integral for general functions, but many treatments define it as the limit of integrals of step functions.

RGV

I'm using Royden, where the integral of a non-negative function is defined as
\int f\, d\,\mu =\sup \{\int \phi \, d\mu :0\le\phi\le f,\,\phi \, simple\}

 

[Kindayr]

So my question is if I'm allowed to say this:
\int f\, d\nu = \sup \{\int \phi\, d\nu :0\le\phi\le f,\, \phi \,simple\} =\sup \{\int \phi g\,d\mu :0\le\phi\le f,\,\phi\,simple\} = \int fg\,d\mu?

 

[Kindayr]

Nevermind, I'll just use Monotone Convergence on a sequence of simple functions.