# Analysis? Set theory?

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.

Note by Ivan Koswara
2 years, 3 months ago

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Given 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 $f(x) = \sum_{i \in \mathbb{Z}} 2^i \mathbf{1}_{S_i}(x).$

- 1 year, 8 months ago