For a subset of real numbers \(S\), let \(\mathbf{1}_S : \mathbb{R} \to \{0, 1\}\) be the **indicator function** of \(S\), defined as \(\mathbf{1}_S(x) = 1\) if \(x \in S\) and \(\mathbf{1}_S(x) = 0\) otherwise.

Prove or disprove: for every real function \(f : \mathbb{R} \to \mathbb{R}\), there exists subsets of real numbers \(A_1, A_2, A_3, \ldots\) and real numbers \(c_1, c_2, c_3, \ldots\) such that

\[\displaystyle\large{ f(x) = \sum_{n=1}^\infty c_n \mathbf{1}_{A_n}(x) }\]

for all real \(x\).

**Clarification**: When I posted this problem, I didn't know the answer. Now, I found the answer, but I find it interesting (like most set theory stuff), so I'll let you to figure it out.

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TopNewestGiven a real number, we can express its binary representation as \(\dots b_2 b_1 b_0.b_{-1} b_{-2} \dots\), where each digit \(b_i\) is 0 or 1.

For an integer \(i\), let \(S_i\) be the set of real numbers \(x\) such that the \(i\)th digit in the binary representation of \(f(x)\) is equal to 1. Then \[f(x) = \sum_{i \in \mathbb{Z}} 2^i \mathbf{1}_{S_i}(x).\] – Jon Haussmann · 11 months ago

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