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Exercise: Let $(X, \mathcal{A}, \mu)$ be a measure space and $(X, \tilde{\mathcal{A}}, \tilde{\mu})$ its completion. Suppose $\tilde{f}: X\to \mathbb{R}$ is $\tilde{\mathcal{A}}-$measurable. Show that there exist $f:X\to \mathbb{R}$ $\mathcal{A}-$measurable and $M\in \mathcal{A}$ with $\mu(X-M) = 0$ such that $\tilde{f}(x) = f(x)$ for all $x\in M$.

My try: Let $r_1, r_2, ...$ be an enumeration of all the rationals. For each natural number $n$, we know $\tilde{f}^{-1}((-\infty, r_n)) = A_n\cup M_n$, where $A_n\in \mathcal{A}$ and $M_n\subseteq M_0^n$ with $\mu(M_0^n) = 0$. The idea is to try and define $f$ in such a way that $f^{-1}((-\infty, r_n))$ is $\mathcal{A}-$measurable for all $n\in \mathbb{N}$ and $f(x) = \tilde{f}(x)$ for all $x$ except maybe those in $\bigcup_{n\in \mathbb{N}} M_0^n$. I can't complete the proof though.

Any suggestions are greatly appreciated.

Asaf Karagila
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Detached Laconian
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1 Answers1

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Let $(r_n)_{n \in \mathbb{N}}$ an enumeration of $\mathbb{Q}$. A function $f:X \to \mathbb{R}$ is $\mathcal{A}$-measurable iff $$\forall n \in \mathbb{N}: f^{-1}((-\infty,r_n)) \in \mathcal{A} \tag{1}$$

Since $\tilde{f}$ is $\tilde{\mathcal{A}}$-measurable, we have $$\tilde{f}^{-1}((-\infty,r_n)) = A_n \cup M_n$$ where $A_n \in \mathcal{A}$, $M_n \subseteq M_0^n \in \mathcal{A}$ such that $\mu(M_0^n)=0$. As a countable union of null sets, the set $$N:= \bigcup_{n \in \mathbb{N}} M_0^n$$ is also a null set and $N \in \mathcal{A}$. Define $f:X \to \mathbb{R}$ by $$f(x) := \begin{cases} \tilde{f}(x) & x \notin N \\ 0 & x \in N \end{cases}$$ Then $[f \not= \tilde{f}] \subseteq N$, thus $f=\tilde{f}$ almost everywhere since $\mu(N)=0$. Moreover, $$f^{-1}((-\infty,r_n)) = \begin{cases} A_n \cap (X \backslash N) & n \in \mathbb{N} \, \text{such that} \, r_n \leq 0 \\ A_n \cup N & n \in \mathbb{N} \, \text{such that} \, r_n > 0 \end{cases} \quad \in \mathcal{A}$$ From $(1)$ we conclude that $f$ is $\mathcal{A}$-measurable.

saz
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  • Why is $$f^{-1}((-\infty,r_n)) = \begin{cases} A_n & n \in \mathbb{N} , \text{such that} , r_n < 0 \ A_n \cup N & n \in \mathbb{N} , \text{such that} , r_n \geq 0 \end{cases} \quad$$ ? – Detached Laconian Mar 16 '13 at 22:01
  • @TuringMachine There was a misprint, I corrected it. Let $n$ such that $r_n<0$. Then $$f^{-1}((-\infty,r_n)) = (A_n \cup M_n) \cap (X \backslash N) = A_n \cap (X \backslash N)$$ since $M_n \cap (X \backslash N) \subseteq N \cap (X \backslash N) = \emptyset$ by definition of $N$. Similar proof works for $n$ such that $r_n \geq 0$. – saz Mar 16 '13 at 22:04
  • These two details don't affect the correctness of your proof but I think $[f \neq \tilde{f}]$ is not necessarily measurable, so one should write instead that $[f \neq \tilde{f}] \subseteq N$ and $\mu(N) = 0$. Also, when $r_n = 0$, $f^{-1}((-\infty, r_n)) = A_n\cap (X\setminus N)$. I was thinking the proof would be more complex, which kind of served as a "mental block" for me to complete it. Thanks a lot for helping me out! – Detached Laconian Mar 16 '13 at 22:40
  • @TuringMachine Yes, of course, you are right. – saz Mar 16 '13 at 22:47