Answer to Question #267131 in Real Analysis for Riya

Question

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.

Answer 1

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,

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

and E_n = 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 ∈ E_n,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 ∈ E_n,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 ∈ E_n 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,

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