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I am trying to understand the proof a theorem in Folland. The theorem says if $(X,M,\mu)$ is a measure space with completion $(X,\bar{M},\bar{\mu})$, and $f$ is $\bar{\mu}$-measurable, then there is a $m$-measurable $g$ such that $f=g$, $\bar{\mu}$-almost everywhere.

He says this is obvious if $f = \chi_E$, the characteristic/indicator function of set $E$. I am not really sure I understood why the theorem is true in this case.

Attempt: If $E$ is not a null set, then we can let $g = \chi_E$. If $E$ is a null set, then $E$ is either in $M$ or $E$ is a subset of a null set in $M$ by definition of completion. If $E$ is in $M$, then let $g = \chi_E$. If not, it is a subset of some null set $S$. Let $g = \chi_S$, so $f = g$ except on $S - E$, which is a null set. Also, we know from earlier in the book that characteristic functions are measurable.

Question: Is my attempt correct? Did I miss out or misunderstood anything?

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If $E$ is a ($\bar{\mu}$)-null set, we can choose $$g := 0$$ because $$\bar{\mu}(f=g) = \bar{\mu}(1_E=0) = \bar{\mu}(X \backslash E)=1,$$ i.e. $f=g \,$ $\tilde{\mu}$-almost everywhere. This simplifies the second part of your proof.

Concerning the first part, i.e. the case that $E \in \bar{M}$ is not a null set: Please note that $g=\chi_E$ is in general not $M$-measurable as $M \subset \bar{M}$. Instead, we have to use a similar argumentation as in your proof: By the definition of the completion, there exists $F \in M$ and a subset $G$ of $(\mu)$-null set such that $$E=F \cup G.$$ Setting $g := \chi_F$ finishes the proof. Let me finally remark that this argumentation also applies to the case that $E$ is a null set (then $F = \emptyset$) and therefore you do not need to consider this case separately.

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  • $\begingroup$ Thanks! +1 and accepted. Although I did not mention it, this is in fact what Folland was referring to. He said it is obvious from the definition of $\bar{M}$. $\endgroup$ – Jean Valjean Dec 27 '13 at 19:36
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If $E\in\overline{\mathcal M}$, then there exist $E_1,E_2\in {\mathcal M}$, such that $$ E_1\subset E\subset E,\,\,\mu(E_2\smallsetminus E_1)=0 \quad \text{and}\quad \overline{\mu}(E)=\mu(E_1)=\mu(E_2). $$ Thus $$ \chi_{E_1}=\chi_{E_2}=\chi_E\quad \text{$\overline{\mu}$-a.e.} $$ For a general $f$, we assume that $f\ge 0$, as if not we can $f$ as $f=f_+-f_-$, where $f_+,f_-$ are its positive and negative parts. Then $f$ can be written as $$ f=\sum_{n\in\mathbb N} a_n\chi_{A_n}, $$ where $a_n>0$ and the $A_n$'s are $\overline{\mu}$-measurable. Choose $B_n\in{\mathcal M}$, s. t. $\chi_{A_n}=\chi_{B_n}$, $\,\,\overline{\mu}$-a.e., and a desired $\mu$-measurable function, $\overline{\mu}$-a.e. equal to $f$, would be $$ g=\sum_{n\in\mathbb N} a_n\chi_{B_n}. $$

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