# Homework Help: Measure Theory Question

1. Feb 20, 2012

### Kindayr

1. The problem statement, all variables and given/known data
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 $\nu$ is a measure and $$\int f\, d \nu =\int fg\,d\mu$$ for all nonnegative measurable functions $f$ on $X$.

3. 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.

2. Feb 20, 2012

### 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.

3. Feb 29, 2012

### 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?

4. Feb 29, 2012

### Kindayr

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

5. Feb 29, 2012

### Ray Vickson

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

6. Feb 29, 2012

### Kindayr

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?

7. Feb 29, 2012

### Ray Vickson

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

8. Feb 29, 2012

### Kindayr

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

9. Feb 29, 2012

### Kindayr

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

10. Mar 1, 2012

### Kindayr

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