Prove that χ ( 0 , 1 ) − χ S is not a limit of increasing step functions.Let A n be sequence of connected bounded subsets (interval) of real numbers. Step function is defined to be the finite linear combination of their charateristic functions. ψ = c 1 χ A 1 + c 2 χ A 2 + . . . + c n χ A n while c k ∈ R for k = 1 , 2 , . . . , n.Let I n be a sequence of open intervals in (0,1) which covers all the rational points in (0,1) and such that ∑ ∫ χ I n ≤ 1 2 . Let S = ⋃ I n and f = χ ( 0 , 1 ) − χ S show that there is no increasng sequence of step function { ψ n } such that lim ψ n ( x ) = f ( x ) almost everywhere. (by means of increasing, ψ n ( x ) ≤ ψ n + 1 ( x ) for all x)I think ψ n = χ ( 0 , 1 ) − ∑ k = 1 n χ I k is what author intended. ∫ ψ n is decreasing and ∫ ψ n ≥ 1 − 1 2 . this shows that lim ∫ ψ n converges. so lim ψ n ( x ) = f ( x ) almost everywhere. However convergence of ψ n doesn't prove the non-existence of increasing step function which converges to f a.e.How can I finish the proof?

Question

Prove that       χ          (      0      ,      1      )        −      χ          S       is not a limit of increasing step functions.Let       A    n   be sequence of connected bounded subsets (interval) of real numbers. Step function is defined to be the finite linear combination of their charateristic functions.  ψ  =      c    1        χ                  A        1              +      c    2        χ                  A        2              +  .  .  .  +      c    n        χ                  A        n            while       c    k    ∈      R   for   k  =  1  ,  2  ,  .  .  .  ,  n.Let             I      n       be a sequence of open intervals in (0,1) which covers all the rational points in (0,1) and such that   ∑  ∫      χ                  I        n              ≤      1    2  . Let   S  =  ⋃      I    n   and   f  =      χ          (      0      ,      1      )        −      χ    S   show that there is no increasng sequence of step function   {      ψ    n    } such that   lim      ψ    n    (  x  )  =  f  (  x  ) almost everywhere. (by means of increasing,       ψ    n    (  x  )  ≤      ψ          n      +      1        (  x  ) for all x)I think       ψ    n    =      χ          (      0      ,      1      )        −      ∑          k      =      1        n        χ                  I        k             is what author intended.   ∫      ψ    n   is decreasing and   ∫      ψ    n    ≥  1  −      1    2  . this shows that   lim  ∫      ψ    n   converges. so   lim      ψ    n    (  x  )  =  f  (  x  ) almost everywhere. However convergence of       ψ    n   doesn&#039;t prove the non-existence of increasing step function which converges to f a.e.How can I finish the proof?

Hassan Watkins · Accepted Answer

Explanation:Suppose such a sequence   {      ψ    n    } exists. Note that       ψ    n    ≤  0 at rational points since   f  ≤  0 at those points. If a step function is non-positive at rational points it is so at all but countable many points. [ The countable many points I am referring to are the end points of teh intervals on which the function is a constant]. It follows that   ∫      ψ    n    ≤  0 and hence   ∫  f  ≤  0 which is clearly a contradiction.

Let An be sequence of connected bounded subsets (interval) of real numbers. Step function is defined to be the finite linear combination of their charateristic functions. psi=c_1 chi_{A_1}+c_2 chi_{A_2}+cdots+c_n chi_{A_n} while c_k in mathbb{R} for k=1,2,...,n

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