Show that the following is a measure on the $\sigma$-algebra of Lebesgue measurable sets

Let $f$ be a nonnegative Lebesgue measurable function on $\mathbb{R}$. For each Lebesgue measurable subset $E$ of $\mathbb{R}$, define $\mu(E) = \int_E f$. Show that $\mu$ is a measure on the $\sigma$-algebra of Lebesgue measurable sets.

Here is what I did:

1. First of all, $\mu$ is nonnegative, as it represents the positive quantity "area".

2. Secondly, $\mu(\emptyset) = \int_{\emptyset} f = a*m(\emptyset) = 0$

3. Lastly, for countable additivity, let $E = \bigcup_{n=1}^{\infty}E_i$, where $E_i \bigcap E_j = \emptyset$ for $i \neq j$.

Then $\mu(E) = \mu(\bigcup_{n=1}^{\infty}E_i) = \int_{\bigcup_{n=1}^{\infty}E_i} f = \sum_{i=1}^{\infty} \int_{E_i}$

By the additivity over disjoint subsets of the domain property of Lebesgue integral.

Does this seem okay?

• For 1. you need to use the non-negativity of f? – Ben Lansdell Nov 21 '16 at 6:07