Knowing that $f -g$ is a simple function I wanna show that $|f – g|$ is again a simple function.
Here is my trial: assuming that $ f = \sum_{i=1}^{n_1} a_i \chi_{A_i}$ and $g = \sum_{j=1}^{n_2} b_j \chi_{B_j}$ I claim that $|f – g|$ is again a simple function because the absolute value of a measurable function is measurable and $|f + (- g)| \leq |f| + |g|$ and $|f|$ and $|g|$ are simple functions because they are less than or equal $\sum_{i=1}^{n_1} |a_i| \chi_{A_i}$ and $\sum_{j=1}^{n_2} |b_j| \chi_{B_j}$ respectively which are simple functions.
Is my argument correct? If no, please help me in correcting it.
Best Answer
The standard way of representing a simple function $\phi$ with distinct values $\{a_1,\ldots a_n\}$ is $$\phi = \sum_{k=1}^n a_k \chi_{A_k}$$ where the $A_k = \{x : \phi(x) = a_k\}$. The $A_k$ are disjoint so that $$|\phi| = \sum_{k=1}^n |a_k| \chi_{A_k}$$ which is also simple.