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

Ruby Briggs 2022-07-20 Answered
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?
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Answers (1)

Hassan Watkins
Answered 2022-07-21 Author has 18 answers
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.
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