# Showing the absolute value of a simple function is a simple function:

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.

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.

• How the absolute value crosses the summation sign without the triangle inequality?
– user965463
Commented Nov 2, 2021 at 1:01
• Focus on the term disjoint. Look at an example when $n=2$ if it helps: if either $a$ or $b$ is zero then $|a+b| = |a| + |b|$. Commented Nov 2, 2021 at 1:04
• Are you saying that the triangle inequality does not apply here because they are disjoint? are you saying that it is applied only when the is intersection?
– user965463
Commented Nov 2, 2021 at 1:08
• The triangle inequality is not used here at all. Commented Nov 2, 2021 at 1:09
• Last comment: if $x_2,x_3,\ldots,x_n$ are all equal to zero, then $$|x_1 + x_2 + \cdots + x_n| = |x_1| = |x_1| + |x_2| + \cdots + |x_n|.$$ Commented Nov 2, 2021 at 1:14

You want to show that if $$f$$ is simple then also $$|f|$$ is simple. You don't need to consider $$f-g$$. Your argument does not hold because $$f(x) \leqslant g(x)$$ where $$g(x)$$ is simple does not imply that $$f$$ is simple. For example $$f(x)= \begin{cases} 0, \text{ if }x < 1\\ -x^2+1, \text{ if }-1\leqslant x \leqslant 1\\ 0, \text{ if } x>1, \end{cases}$$ is not simple but it is measurable and it is less or equal to the simple function $$\mathcal{X}_{[-1,1]}$$.

Let $$f$$ be a simple function $$\displaystyle\sum_{i=1}^{n} a_i \mathcal{X}_{A_i}$$ where $$\mathcal{X}_{A_i}$$ is the indicator function of $$A_i$$. Let $$f^{+}$$ be the positive part of $$f$$ and let $$f^{-}$$ be the negative part of $$f$$. (So that $$f=f^{+}-f^{-}$$).Note that $$f^{+}$$ and $$f^{-}$$ are simple functions. Since the support of $$f^{+}$$ and $$f^{-}$$ are disjoint we simply have $$|f|=f^{+}+f^{-}$$ and this is the sum of two simple functions, which is simple.

• What do you mean by the support of $f^+$?
– user965463
Commented Nov 2, 2021 at 0:59
• @Brain By support I mean where the function is not 0. (See en.wikipedia.org/wiki/Support_(mathematics))
– Lisa
Commented Nov 2, 2021 at 8:39

Let $$(X, \mathcal{F})$$ be a measurable space. A function $$\phi : X \to \mathbb{C}$$ is simple if and only if $$\phi$$ is measurable and has finite range $$\phi(X)$$. If $$\phi$$ is simple, then $$|\phi|$$ is measurable and has finite range $$|\phi(X)|$$, so $$|\phi|$$ is simple.