Let $\mu$ be a measure on $(X, \mathcal{A})$ and a measurable function $f:X \to \mathbb{R}, \ f \geq 0$.

Define $\mu_f(E): \mathcal{A} \to \mathbb{R}, \ \mu_f(E):=\int_E f \ d\mu$ for $E \in \mathcal{A}$.

How to prove that $\mu_f$ is a measure on the sigma-algebra $\mathcal{A}$?

I tried it with:

$\mu_f(\emptyset)=\int_\emptyset f \ d\mu = 0$.

I'm not sure if this is right.

For the countable additivity I don't know how to show that

$\mu_f(\cup^{i=1}_{\infty}E_i)=\sum_{i \in I}{\mu_f(E_i)}$.


1 Answer 1



Further be aware that we always have $\int\sum_{i=1}^{\infty}g_i\;d\mu=\sum_{i=1}^{\infty}\int g_i\;d\mu$ if the $g_i$ are measurable and nonnegative.

By disjoint and measurable $E_i$ moreover we have $\mathbf1_{\bigcup_{i=1}^{\infty}E_i}=\sum_{i=1}^{\infty}\mathbf1_{E_i}$ so that:



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .