# Unique representation of simple function

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?

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

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 $$\chi_A$$.