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How do I prove this proposition from Royden's Real Analysis:

If $\mu$ is a complete measure and $f$ is a measurable function, then $f=g$ almost everywhere implies $g$ is measurable.

In proving this proposition, what differs from the proof of a proposition from the first chapters stating:

If $f$ is a measurable function $f=g$ almost everywhere then $g$ is measurable.

In particular, what has to be modified in the following proof:

Take $E=\lbrace x \in X | f(x) \neq g(x) \rbrace,$ which is measurable and has measure $0$. For a measurable set $A$ in the range of $g$, we show that the set $Y=g^{-1}(A)$ is measurable. Now, $Y \cap E$ has is measurable with measure $0$. Since $Y \setminus E = f^{-1}(A) \setminus E$ is a difference of two measurable sets, we are done.

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You don't know that $E$ is measurable. All you know is that $E$ is a null-set, which by definition means that $E\subseteq F$, for some measurable null-set $F$, i.e. $F$ is measurable and $\mu(F)=0$. As Ilya points out: you cannot conclude that $Y\cap E$ is a measurable null-set, only that it is a null-set. – Stefan Hansen Jan 18 '13 at 9:55
Now, I'm lost on how to go about proving this. – user58191 Jan 18 '13 at 10:16
@StefanHansen Since the measure $\mu$ is complete, any subset of a null set is measurable; in particular, $E$ is measurable. Or am I missing something obvious? – saz Jan 30 '15 at 15:40
@saz: Yeah, you're right. I don't know what I was thinking of at the time. – Stefan Hansen Jan 30 '15 at 15:54

Please, check assumptions of the proposition from the first chapters. When you are saying that $Y\cap E$ is measurable - how do you know this? You are not given that $Y$ is measurable, you have to prove it. But in case the measure is complete, by the definition of completeness if follows that $$ Y\cap E \subset E\text{ and }\mu(E) = 0\quad \Rightarrow \quad Y\cap E \text{ is measurable}. $$ Without completeness you can only conclude that $Y\cap E$ is $\mu$-null which does not imply measurability in general.

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I wasn't claiming $Y$ was measurable right away, I thought it's implied by the last two sentences. I guess $Y \cap E$ being measurable comes from the completeness property, is that the only gap from my attempt at proof? – user58191 Jan 18 '13 at 10:21
@user58191: I rather meant, that the measurabiltiy of $Y\cap E$ could be concluded either from measurability of $Y$ and the one of $E$, or using the fact that it is $\mu$-null and $\mu$ is complete. What I wrote is that you can't go the first way. Anyway, please tell me if you are still confused - I'll elaborate on unclear moments – Ilya Jan 18 '13 at 10:24
It's a little clearer but I'm still a bit confused. So do I just add in the bit about $\mu$-completeness and $\mu$-nullity implying $Y \cap E$ is measurable (as above)? – user58191 Jan 18 '13 at 10:32
I meant, is that the only gap in what I have written above? – user58191 Jan 18 '13 at 10:53
Thanks! I think I get it already! – user58191 Jan 18 '13 at 12:08

To fix notation: Suppose that $(X,\mathcal{A},\mu)$ is a complete measure space, $(Y,\mathcal{B})$ is a measure space and that $f:X \to Y$ is measurable. We have to show that

$$g^{-1}(B) \in \mathcal{A}$$

for all $B \in \mathcal{B}$. Now

$$\begin{align*} g^{-1}(B) &= \{x; g(x) \in B \} \\ &= \underbrace{\{x; g(x) \in B, f(x)=g(x)\}}_{=:A} \cup \underbrace{\{x; g(x) \in B, g(x) \neq f(x)\}}_{=:N}. \end{align*}$$

By definition, we have

$$N \subseteq \{x; f(x) \neq g(x)\}.$$

Since $f=g$ almost everywhere, this shows that $N$ is a subset of a nullset; hence measurable as $\mu$ is complete. Moreover,

$$\begin{align*} A &= \{x; g(x) \in B, f(x)=g(x)\} \\ &= \{x; f(x) \in B, f(x)=g(x)\} \\ &= \{x; f(x) \in B\} \cap \{x; f(x)=g(x)\} \end{align*}$$

is measurable as the intersection of two measurable sets (the first set at the right-hand side is measurable because $f$ is measurable and for the second one we note that

$$\{x; f(x)=g(x)\} = X \setminus \{x; f(x) \neq g(x)\}$$

is the complement of a measurable set and therefore measurable.)

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I'm sorry, I should have mentioned that I'm looking for a proof of the more general case, $f:X\to Y$, where $Y$ is not necessarily the real line. (Just for the record, I found this elegant proof , problem 1a, for the case $Y=\mathbb R$) – fonini Jan 30 '15 at 20:18
@fonini The proof works exactly the same way for the more general case. See my edited answer... – saz Jan 30 '15 at 20:19
Well, that was pretty simple indeed. Thanks :) – fonini Jan 30 '15 at 20:26

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