$\nu(A) = \int_A f\,d\mu$ is a measure? Let $(X,\mathcal A, \mu)$ be a measure space and let $f$ be nonnegative and integrable. Define $\nu$ on $A$ by$$\nu(A) = \int_A f\,d\mu.$$How do I see that $\nu$ is a measure?
 A: Yes, it is (even if $f$ is not integrable), as long as $f$ is measurable. To see this, note that


*

*$\nu(A)=\int_A f\,\mathrm d\mu\geq 0$, given that $f\geq 0$;

*$\nu(\varnothing)=\int_{\varnothing} f\,\mathrm d\mu= 0$;

*if $(A_n)_{n\in\mathbb N}$ is a sequence of disjoint measurable sets, then $$\nu\left(\bigcup_{n\in\mathbb N} A_n\right)=\int_{\bigcup_{n\in\mathbb N}A_n}f\,\mathrm d\mu=\sum_{n\in\mathbb N}{\int_{A_n}}f\,\mathrm d\mu=\sum_{n\in\mathbb N}\nu(A_n).$$


In addition, it $f$ is integrable, then $\nu$ will be a finite measure: $$\nu(X)=\int_X f\,\mathrm d\mu<\infty.$$
A: If $\{A_n\}$ is a sequence of disjoint sets in $\mathcal A$ then by monotone convergence we have
\begin{align}
\nu\left(\bigcup_{n=1}^\infty A_n\right) &= \int_{\bigcup_{n=1}^\infty A_n} f\,\mathsf d\mu\\
&=\int f\mathsf 1_{\bigcup_{n=1}^\infty A_n}\,\mathsf d\mu \\
&=\int f\sum_{n=1}^\infty \mathsf 1_{A_n}\,\mathsf d\mu\\
&= \sum_{n=1}^\infty \int f\mathsf 1_{A_n}\,\mathsf d\mu\\
&= \sum_{n=1}^\infty \int_{A_n} f\,\mathsf d\mu\\
&= \sum_{n=1}^\infty \nu(A_n),
\end{align}
so $\nu$ is countably additive and therefore a measure.
