# $f$ measurable with $f=g$ a.e. then $g$ measurable

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
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.
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
@user58191: yes, the only part missing is the measurability of $Y\cap E$ which is justified due to the completeness of $\mu$ –  Ilya Jan 18 '13 at 11:37