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Suppose $(X, \mathcal{M}, \mu)$ is a measure space. Assume $f: X\to\overline{\mathbb{R}}$ and $g=X\to\overline{\mathbb{R}}$ are measurable maps. Here $\overline{\mathbb{R}}$ denotes the set of extended real numbers. My question is:

If $f=g$ almost everywhere, does it follow that $|f|=|g|$ almost everywhere?

I know the answer is "Yes" if $X$ is a complete measure space: If $f=g$ a.e. then $E=\{x\in X: f(x)\neq g(x)\}$ is a null set, i.e. $\mu(E)=0$. It is clear that $$ F=\{x\in X : |f(x)|\neq |g(x)|\}\subseteq E $$ Since $X$ is complete, all subsets of null sets are in $\mathcal{M}$, and so $\mu(F)=0$, and $|f|=|g|$ a.e.

What happens when $X$ is not complete?

Thanks for your time :)

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  • $\begingroup$ I think $F$ has to be in $M$ anyway, because $|f|$ and $|g|$ are also measurable functions, no? Since $|\cdot|$ is continuous. Then we have $F \subset E$, $\mu(E) = 0$ and $F \in M$ and thus $\mu(F) = 0$. $\endgroup$ – user38355 Nov 4 '13 at 23:38
  • $\begingroup$ @brom: Thanks! I understand why $|f|$ and $|g|$ are measurable. And $F\in\mathcal{M}$ because it is equal to $(|f|^{-1}(\overline{\mathbb{R}})\cap |g|^{-1}(\overline{\mathbb{R}}))^{c}$, right? $\endgroup$ – Prism Nov 4 '13 at 23:44
  • $\begingroup$ Almost. $|f|$ and $|g|$ are measurable and $F \in M$ because $F = ((|f|-|g|)^{-1}(\{0\}))^c$. I don't think $F = |f|^{-1}(\overline{\mathbb{R}}) \cap |g|^{-1}(\overline{\mathbb{R}})$ is true; in fact this last set is all of $X$. $\endgroup$ – user38355 Nov 4 '13 at 23:58
  • $\begingroup$ @brom: Of course... I don't know what I was thinking... :/ $\endgroup$ – Prism Nov 5 '13 at 0:07
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Since $f,g$ are measurable, so are $f-g$ and $f+g$. Let $$ E=\{x\in X:\ f(x)=g(x)\}, \ \ E'=\{x\in X:\ f(x)=-g(x)\}. $$ If $F$ is as in the question, the set where $|f|\neq|g|$, then we have $F^{c}=E\cup E'$. And then $F$ is measurable, because both $E$ and $E'$ are: $$ E=(f-g)^{-1}(\{0\}),\ \ E'=(f+g)^{-1}(\{0\}). $$

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  • $\begingroup$ Thank you. This is very nice. I like how it only uses the fact that $f+g$ and $f-g$ are measurable. $\endgroup$ – Prism Nov 4 '13 at 23:55
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Note that since $f$ and $g$ are measurable, then so are $|f|$ and $|g|$ by continuity of $|\cdot|$, and hence, $h=|f|-|g|$ is measurable. Noting that $$F=\left(h^{-1}\bigl(\{0\}\bigr)\right)^c,$$ we readily have $F\in\mathcal M$. By nonnegativity and monotonicity of measure, since $F\subseteq E$ and $\mu(E)=0,$ then $\mu(F)=0,$ and so $|f|=|g|$ a.e.

There is no need for $(X,\mathcal M,\mu)$ to be a complete measure space.

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  • $\begingroup$ Efficient solution! Thanks. $\endgroup$ – Prism Nov 4 '13 at 23:56

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