Let $f\colon X\to\overline{\mathbb{R}}_{\geq 0}$ be a measurable function which only takes a countable number of values. Show that $$ \int f\, d\mu=\sum\limits_{y\in f(X)}y\cdot\mu\left(f^{-1}(y)\right). $$
Here is my idea resp. proof:
Consider $(X,\mathfrak{U},\mu)$ and $f$ as mentioned above. I call the values of $f$ as $y_1,y_2,y_3,...$ (resp. finite values). Now set $$ X_{y_i}:=f^{-1}(\left\{y_i\right\}). $$ Then $X$ is the disjunct union of these sets, i.e. $$ X=\bigcup\limits_{y_i\in f(X)}X_{y_i}, X_{y_i}\cap X_{y_j}=\emptyset~\forall~i\neq j. $$ Because for the Lebesgue integral it is$\int\limits_{A\cup B}f(x)\, d\mu=\int\limits_A f(x)\, d\mu+\int\limits_B f(x)\, d\mu$ for disjunct sets $A,B\in\mathfrak{U}$, it is $$ \int_X f(x)\, d\mu=\sum\limits_{y_i\in f(X)}\int\limits_{X_{y_i}}f(x)\, d\mu=\sum\limits_{y_i\in f(X)}\int_X\chi_{X_{y_i}}(x)f(x)\, d\mu $$ The functions $$ \chi_{X_{y_i}}(x)f(x)=\begin{cases}y_i, & x\in X_{y_i}\\0, &\mbox{ elsewhere}\end{cases} $$ are simple functions and for those functions the Lebesgue integral is defined as $$ \int\limits_X\chi_{X_{y_i}}(x)f(x)\, d\mu=y_i\cdot \mu\left(f^{-1}(\left\{y_i\right\})\right). $$ So it is $$ \sum\limits_{y_i\in f(X)}\int_X\chi_{X_{y_i}}(x)f(x)\, d\mu=\sum\limits_{y_i\in f(X)}y_i\cdot\mu\left(f^{-1}(\left\{y_i\right\})\right) $$
Finished!
I would like to know if my proof is correct.
With regards!