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 :)

  • $\begingroup$ 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

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: $$ E=(f-g)^{-1}(\{0\}),\ \ E'=(f+g)^{-1}(\{0\}). $$

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.