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Claim: Consider a simple function $f : X \to \mathbb{R}$ where $(X,M,\mu)$ is a measure space. If we represent $f$ as $\sum_{i=1}^{n} a_{i} \chi_{A_i}$, with $A_i \cap A_j = \varnothing$ and $a_i \ne a_j$ for $i \ne j$, then is this representation of the simple function is unique.

I intuitively feel it should be true, but I'm not able to prove it.

My attempt at proof: Suppose $f = \sum_{i=1}^{n} a_{i} \chi_{A_i} = \sum_{j=1}^{m} b_{j} \chi_{B_j}$. Then we need to prove the following: $n=m$, $a_i = b_{p_i}$ and $A_i = B_{p_i}$, where $p_1,p_2,\ldots,p_n \in \{1,2,\ldots,n\}$ for all $i=1,2,\ldots,n$ and $p_i \ne p_j$ if $i \ne j$. However, I'm stuck at this point and do not know how to proceed.

I have two questions:

  1. Could you suggest how do I complete the proof or some other method to prove the claim, or a counterexample to the claim?

  2. If the claim is true, does it hold without can $a_i \ne a_j$ be removed from the hypothesis, and then established as a part of the proof?

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  • $\begingroup$ Second, note that $\chi_A=\chi_A+0\chi_B$; to get uniqueness up to reordering you need to assume $a_j\ne0$ and $b_j\ne0$. $\endgroup$ Commented Mar 9, 2021 at 22:10
  • $\begingroup$ @DavidC.Ullrich I did not understand what $\chi_A+$ means. And yes I understand what you mean by uniqueness upto reordering. I have edited the question to reflect that. $\endgroup$ Commented Mar 9, 2021 at 22:10
  • $\begingroup$ I didn't write $\chi_A+$. I pointed out that $\chi_A=\chi_A+0\chi_B$. $\endgroup$ Commented Mar 9, 2021 at 22:12
  • $\begingroup$ In the first comment you have used $\chi_A+ = -\chi_B$, and I did not understand what this means. $\endgroup$ Commented Mar 9, 2021 at 22:14
  • $\begingroup$ must have been a typo - the first comment is gone. $\endgroup$ Commented Mar 9, 2021 at 22:15

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Of course it's easy to construct examples showing that you need to assume $$a_j\ne0$$ and $$b_k\ne0.$$ IF you assume that:

HINT Show that $f(X)\setminus\{0\}=\{a_1,\dots,a_n\}$.

Hence $$\{b_1,\dots,b_m\}=\{a_1,\dots,a_n\},$$ and now since the $a_j$ are distinct and the $b_k$ are distinct you're on your way...

Note another way to fix it is to assume that $\bigcup A_j=X$. (If you assume that you had better not assume $a_j\ne0$.) The stuff at the start maybe comes out more elegant with the second version, but once we've moved on the second seems silly, requiring $\chi_A+0\chi_{X\setminus A}$ in place of a simple $\chi_A$.

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