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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.

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3 Answers 3

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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.

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

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

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