Integral with respect to an infinite sum of measures

Question

Let $(\Omega,\mathcal{F})$ be a measurable space and $f : \Omega \rightarrow [0 , \infty]$ be $\mathcal{F}$-measurable. For a fixed $\omega \in \Omega$ define $$ \delta_\omega(A)=\begin{cases} 1 & \omega \in A\\ 0 & \omega \not\in A \end{cases}$$

Show that

(a) $$\int_{\Omega} f \ d \delta_\omega = f(\omega) $$ (b) If $\mu=\sum_{j=1}^\infty c_j \delta_{\omega_j}$ with $c_j \geq 0$ and fixed $\omega_j \in \Omega$, then

$$\int_{\Omega} f \ d \mu = \sum_{j=1}^\infty c_j f(\omega_j)$$

I know that $\mu$ is indeed a measure as an infinite sum of positive measures. I also managed to show (a) by taking an increasing sequence of simple functions converging pointwise to $f$ and using the monotnone convergance theorem.

Can someone show (b), as detailed as possible, with strict justification of every step that deals with limits or infinite sums?

Answer 1

Taking the sequence of simple functions $g_n:=\sum_{j=1}^nf(\omega_j)\chi_{\{\omega_j\}}$ we have $$ \int_\Omega g_nd\mu=\sum_{j=1}^nf(\omega_j)\int_\Omega\chi_{\{\omega_j\}}d\mu=\sum_{j=1}^nf(\omega_j)\mu(\{\omega_j\})=\sum_{j=1}^nf(\omega_j)\underbrace{\left(\sum_{k=1}^\infty c_k\delta_{\omega_k}(\{\omega_j\})\right)}_{c_j},$$ which, denoting $A=\{\omega_j:j\geq1\}$ and applying the monotone convergence theorem, lets us conclude $$ \int_\Omega fd\mu=\int_\Omega\chi_Afd\mu=\int_\Omega\left(\lim_{n\to\infty}g_n\right)d\mu=\lim_{n\to\infty}\int_\Omega g_nd\mu=\sum_{j=1}^\infty c_jf(\omega_j). $$