Answer to Question #267131 in Real Analysis for Riya

Question #267131

If function f is defined on a sigma- finite measure space.we may choose the function pi n so that each vanishes outside a set of a finite measure.

1
Expert's answer
2022-02-21T18:55:51-0500

For each k ≥ 1, we partition the interval [0,∞] into disjoint interval as follows:

[0,∞] = [0, 1) "\\bigcup" [1, 2)"\\bigcup" · · · [n − 1, n) "\\bigcup" [n,∞]


"\\bigcup^{n2^n}_{ i=1}" "[\\frac{i-1}{2^n},\\frac{i}{2^n})\\bigcup [n,\\infin)"


Define, for each 1 ≤ i ≤ n2n,

Eni = f −1"([\\frac{i-1}{2^n},\\frac{i}{2^n}))"


and En = f −1 ([n, ∞]). Define the function "\\pi_n" (x) as follows:


"\\displaystyle{\\sum_{i=1}^{n2^n}}\\frac{i-1}{2^n}\\chi _{E_{n,i}}+n\\chi_{E_n}"


If f(x) < ∞, then there exists a positive integer n such that f(x) < n. Since x ∈ En,i for some

1 ≤ i ≤n2n , we have

0 ≤ f(x) −"\\pi_n"(x) ≤ "\\frac{i}{2^n}-\\frac{i-1}{2^n}=\\frac{1}{2^n}"


If f(x) = ∞, then "\\pi_n"(x) = n for every n. Thus, for each x ∈ E,

"\\displaystyle{\\lim_{n\\to\\infin}}\\pi_n(x)=f(x)"


This gives us a sequence "\\pi_n"(x) of simple functions on E converging to f(x).

We now show that "\\pi_n"(x) ≤ "\\pi_{n+1}"(x) for every integer x ∈ E. To do so, notice that for each

1 ≤ i ≤ "n2^n" ,


"E_{n+1,2i-1}\\bigcup E_{n+1,2i}=f^{-1}([\\frac{2i-2}{2^{n+1}},\\frac{2i-1}{2^{n+1}}))\\bigcup f^{-1}([\\frac{2i-1}{2^{n+1}},\\frac{2i}{2^{n+1}}))=f^{-1}([\\frac{i-1}{2^{n}},\\frac{i}{2^{n}}))=E_{ni}"


If x ∈ En,i for some 1 ≤ i ≤ "n2^n" , then

"\\pi_n(x)=\\frac{i-1}{2^n}"

On the other hand,

"\\pi_{n+1}(x)=\\frac{2i-2}{2^n}" or "\\frac{2i-1}{2^n}"


both of which are

"\\ge \\frac{i-1}{2^n}"


If x ∈ En then "\\pi_n"(x) = n. But,

"f^{-1}([n,\\infin])=f^{-1}(n,n+1)\\bigcup f^{-1}([n+1,\\infin])"

If "x\\isin =f^{-1}(n,n+1)", then

"\\pi_{n+1}(x)=\\displaystyle{\\sum_{i=n2^{n+1}+1}^{(n+1)2^{n+1}}}\\frac{i-1}{2^{n+1}}\\chi_{E_{n,i}}\\ge \\frac{n2^{n+1}}{2^{n+1}}=\\pi_n(x)"


If x ∈ f−1([k + 1,∞]), then "\\pi_{n+1}"(x) = k + 1 > k = "\\pi_{n}" . thus, we have shown that for every x ∈ E,

"\\pi_{n+1}(x)\\ge \\pi_{n}(x)"

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