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**1. The problem statement**

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

[tex]\mu_f (E) = \int_E fd\mu [/tex]

Show that [tex]\mu_f[/tex] 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 [tex]\phi[/tex] be a measurable step function for f and denote [tex]A = \cup_{j=1}^{\infty} E_j [/tex]

then we have that

[tex] \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 [/tex]

Now let [tex]\phi_n[/tex] be a monotonically increasing sequence of measurable step functions that converge to f pointwise.

We can now say that:

[tex]\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)[/tex]

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?