Measurability of piecewise constant function

Question

If $(X, \mathbb A, m)$ is a measurable space and if we have a function $f: (X, \mathbb A) \to (\mathbb R, B(\mathbb R))$ which is measurable, non-negative and piecewise constant, why does - in this case - hold that $$\int f dm = \sum_{a \in f(X)} a \cdot m( \{ x \in X: f(x) = a \}) $$

Answer 1

If the image of $f$ is finite then the equation is clear since we can write $f$ as a simple function and use the definition of the integral. Otherwise, if the image of $f$ is countable, we can order the image set as $(y_1, y_2, \dotsc,)$ in increasing order (we want to avoid uncountable sums) and write $f$ as the limit of the increasing sequence of simple functions $\sum_{j=1}^k y_j I(f(x)=y_j)$ from which the desired result is clear by the MCT theorem.