# Question on measure theory

1. Dec 28, 2006

### Pietjuh

1. The problem statement

Let (X,M,$$\mu$$) be a measure space and let $$f:X \to [0,\infty]$$ be a measurable function. Now define for $$E\in M$$ the following function:

$$\mu_f (E) = \int_E fd\mu$$

Show that $$\mu_f$$ is a measure on M.

3. The attempt at a solution
I will skip the part where I have to show that the measure of the empty set is zero.

Let $$\phi$$ be a measurable step function for f and denote $$A = \cup_{j=1}^{\infty} E_j$$

then we have that
$$\int_A \phi d\mu = \sum_{i=1}^n a_i \mu ( A \cap E_i) = \sum_{i=1}^n a_i \mu ( \cup_{j=1}^{\infty} (E_i \cap E_j) ) = \sum_{i=1}^n \sum_{j=1}^{\infty} a_i \mu (E_i \cap E_j ) = \sum_{j=1}^{\infty} \int_{E_j} \phi d\mu$$

Now let $$\phi_n$$ be a monotonically increasing sequence of measurable step functions that converge to f pointwise.
We can now say that:

$$\mu_f (A) = \int_A fd\mu = \lim_{n\to\infty} \int_A \phi_n d\mu = \lim_{n\to\infty} \sum_{j=1}^{\infty}\int_{E_j} \phi_n d\mu = \sum_{j=1}^{\infty}\lim_{n\to\infty}\int_{E_j} \phi_n d\mu = \sum_{j=1}^{\infty} \mu_f (E_j)$$

The trouble with my solution is in this last step I think. I'm not sure if you can always find suitable step functions that converge monotonically to f, and I'm also not sure if I'm allowed to exchange the limit with the infinite sum in this step.

Can someone point me out if what I'm doing here is right or wrong?

2. Dec 28, 2006

### matt grime

Why are you constructing step functions that converge to f?

If you want to show something is a measure, show it satisfies the (two, right?) axioms of a measure. The measureable sets are even given to you.

3. Dec 28, 2006

### Pietjuh

Yes I have to show that 1. the measure of the empty set is zero, and 2. that the measure of a countable union of measureable sets is the sum of the measures of the individual measurable sets.

So in this case I need to show that the integral over a domain that is a countable union of measurable sets, can be written as a sum of integrals over the individual measurable sets. And the way to proceed with this, was my thought, is to use the measurable step functions, because integrals are defined by means of measurable step functions!

4. Dec 28, 2006

### matt grime

It is unnecessary, and doesn't explain why you're making things converge to f, which is known integrable function with respect to mu. And that is the key - you already have something you know to be a measure, mu, that you're actually integrating with respect to. You also omitted the fact that the sets need to be *disjoint*.

The measure of a set is the integral of the indicator function over that set. You're given the measurable sets - they are the same as for mu. The measure of any set A, is then just the integral of f over A. This is trivially a measure since A and f are measurable with respect to mu.

5. Dec 28, 2006

### Pietjuh

I'm still not entirely convinced by your argument, especially this sentence: "The measure of any set A, is then just the integral of f over A"

I understand that the measure of a set is just the integral of the indicator function. But I still don't understand why it's trivially true that if the measurable sets are the same and that if f is measurable that the integral over f is trivially a measure.

call me stupid if you want, but i don't get it :(

6. Dec 28, 2006

### matt grime

f is a measurable function with respect to M and mu. So by definition the integral of f over any mu measurable set, i.e. something in M, exists, and satisfies the countable additivity condition, and the integral of f over an empty set is zero. Right? Well, that proves f*mu is a measure on M, doesn't' it?