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Approximating simple function in $L_1$ with countable base

  1. Nov 8, 2014 #1
    Let ##f:X\to \mathbb{Q}+i\mathbb{Q}\subset\mathbb{C}##, ##f\in L_1(X,\mu)## be a Lebesgue-summable function taking only finitely many values ##y_1,\ldots,y_n\in \mathbb{Q}+i\mathbb{Q}## on the sets ##E_1,\ldots,E_n## such that ##\bigcup_{i=1}^nE_i=X##, ##\forall i\ne j\quad E_i\cap E_j=\emptyset##. Space ##X## is endowed with measure ##\mu##, which has a http://librarum.org/book/10022/196 [Broken] that is a ring of sets.

    By definition of countable base, for any ##E_k, k=1,\ldots,n## and any ##\varepsilon>0## there is a set of the basis ##A_k## such that ##\mu(E_k\triangle A_k)<\varepsilon##. Let us define ##A_k':=A_k\setminus\bigcup_{i<k} A_i## and the function
    ##f^\ast(x)= \begin{cases} y_k & \text{if } x\in A_k' \\ 0 & \text{if } x\in X\setminus\bigcup_{i=1}^n A_i' \end{cases}##

    Then, Kolmogorov and Fomin's Introductory Real Analysis http://librarum.org/book/10022/196 [Broken] that, "clearly", for ##\varepsilon## small enough, ##\mu\{x\in X:f(x)\ne f^\ast (x)\}## can be made arbitrarily small.

    Could anybody show how to prove that ##\mu\{x\in X:f(x)\ne f^\ast (x)\}<\delta## for some choice of ##A_k##'s ?

    I notice that, since ##(\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k')=(\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k)\subset\bigcup_{k=1}^n(E_k\triangle A_k)##, the inequality ##\mu((\bigcup_{k=1}^n E_k)\triangle(\bigcup_{k=1}^n A_k'))\leq\sum_{k=1}^n\mu(E_k\triangle A_k)<n\varepsilon ## holds. We also have that ##f^\ast(x)\ne f(x)\Rightarrow x\in X\setminus\bigcup_{k=1}^n(A_k'\cap E_k)=\bigcap_{k=1}^n (A'^c_k\cup E_k^c) ## where I use the notation ##S^c:=X\setminus S##. An inclusion of ##\bigcap_{k=1}^n (A'^c_k\cup E_k^c) \subset M## such that ##\mu(M)## is arbitrarily small would prove the desired result, but I am not sure we can find one...
    Thanks such that ##\mu##(my thanks)##=+\infty##! ;)
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 13, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 14, 2014 #3
    Thank you so much! No, I have come to no new conclusion...
     
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