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Math Help - Step Functions

  1. #1
    Senior Member slevvio's Avatar
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    Step Functions

    Let S = \displaystyle\bigcup_{r=1}^n E_n be the union of distinct intervals E_1, \ldots E_n in \mathbb{R} i.e. none of the intervals are the same

    I am trying to show that \chi_S is a step function \implies E_1,\ldots E_n are disjoint. I am a bit stumped.

    So assume \chi_S = \lambda_1 \chi_{E_1} + \ldots + \lambda_n \chi_{E_n}.

    If there is only one non-empty intersection of two intervals, it is easy to show that this is impossible, but I am unable to prove it in the general case, i.e. when there is an arbitrary number of collections of intervals with non-empty intersection, has anyone got any tips? Thanks very much
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    Let S = \displaystyle\bigcup_{r=1}^n E_n be the union of distinct intervals E_1, \ldots E_n in \mathbb{R} i.e. none of the intervals are the same

    I am trying to show that \chi_S is a step function \implies E_1,\ldots E_n are disjoint. I am a bit stumped.

    So assume \chi_S = \lambda_1 \chi_{E_1} + \ldots + \lambda_n \chi_{E_n}.

    If there is only one non-empty intersection of two intervals, it is easy to show that this is impossible, but I am unable to prove it in the general case, i.e. when there is an arbitrary number of collections of intervals with non-empty intersection, has anyone got any tips? Thanks very much
    I assume chi is the indicator function, allow me to use the bolded one. Assume that \displaystyle \mathbf{1}_S=\sum_{j=1}^{m}\alpha_j \mathbf{1}_{A_j} where \{A_j\}_{j\in[m]} is an interval partition of S. Assume that \{E_k\}_{k\in[n]} is not a pairwise-disjoint set. Then, we may assume (with the possibility of relabeling) that x\in E_1\cap E_2. Show that this implies that \mathbf{1}_S(x)\geqslant 2.

    If this was unhelpful, in the sense that I didn't really say anything substantiative, then just tell me and I'll try to give a better hint.
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  3. #3
    Senior Member slevvio's Avatar
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    Well \mathbf{1}_S (x)=1 on E_1, E_2, and  E_1 \cap E_2. So \mathbf{1}_S (x) = \lambda_1 + \lambda_2 + other possible terms = 1 on E_1 \cap E_2. I just don't see why this has to be \ge 2.

    It is if E_1 and E_2 are the only sets that intersect, but if not you get random other sets coming and adding different values of positive or negative lambdas that alter the whole sum.
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