Lebesgue integral of $\int_\Omega fd\mu$ where $\mu(A)=\sum\limits_{k=1}^{n}\alpha_k\cdot\delta_{x_k}$

Question

I have an example in my lecture notes where the following is stated:

Let $\delta_{x_k}$ be the Dirac measure at $x$. Note that for $\mu(A)=\sum\limits_{k=1}^{n}\alpha_k\cdot\delta_{x_k}$ the integral is the sum

$\hspace{5cm}$ $\int_\Omega fd\mu=\sum\limits_{k=1}^{n}\alpha_k\cdot f(x_k)\hspace{2cm}(1.1)$

In particular $\int_\Omega fd\delta_x=f(x)$.

They don't mention any particular measure space here which is anyways irrelevant I think. My question now is how they derived the sum expression of the integral in $(1.1)$ and then afterwards the integral w.r.t to $\delta_x$

Answer 1

$\int f d\mu=\sum\limits_{k=1}^{n} \int_{\{x\}} fd\mu+\int_A fd\mu$ where $A=\Omega \setminus \{x_1,x_2,..,x_n\}$. The last term is $0$ because $\mu(A)=0$. On $\{x\}$ the function $f$ has the consatnt value $f(x_k)$. Hence $\int_{\{x\}} fd\mu=f(k) \mu (\{x_k\})=f(k)\alpha_k$.

Answer 2

I believe this is used:

If $v_1,v_2$ are two measures on the same measurable space space and $f$ measurable with respect to both measures, and $v=v_1+v_2$ then $\int fdv=\int f dv_1+\int fdv_2$

This can be proved with approximation by simple functions and then proceed by induction.