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So sorry to bother y'all, I'm on a bit of a deadline and am not an expert on this and can't find the reference.

So it turns out that a measurable function defined on a subset $U$ of a measurable space $X$ can be extended to a measurable function on the whole space by defining it to be $0$ on $X\setminus U$. I needed the reference for this, anyone? Thanks

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1 Answer 1

up vote 4 down vote accepted

I'll use $\overline{\mathbb{R}}$ to denote the extended real numbers, i.e. $[-\infty,\infty]$. But the argument works just as well with plain old $\mathbb{R}$, or even any measurable subset of $\mathbb{R}$ containing $0$, or even any measurable space $Y$ (with some $y\in Y$ replacing $0$).

Let $X$ be a measurable space with $\sigma$-algebra $\mathcal{M}$, and let $U\subseteq X$ be a measurable set (i.e. $U\in\mathcal{M}$), made into a measurable space with the $\sigma$-algebra $\mathcal{N}=\{A\cap U\mid A\in\mathcal{M}\}$. Note that $\mathcal{N}\subseteq\mathcal{M}$, because the intersection of any two sets in $\mathcal{M}$ is in $\mathcal{M}$.

Let $f:U\to\overline{\mathbb{R}}$ be a measurable function, and define $F:X\to\overline{\mathbb{R}}$ by $$F(x)=\begin{cases} f(x)\quad\,\text{ if }x\in U\\0\quad\quad\quad\text{ if }x\notin U.\end{cases}$$ $F$ is measurable if and only if, for any measurable subset $S\subseteq\overline{\mathbb{R}}$, the set $F^{-1}(S)$ is a measurable subset of $X$, i.e. $F^{-1}(S)\in\mathcal{M}$.

If $0\notin S$, then $F^{-1}(S)=f^{-1}(S)$. Because $f$ is measurable, we have that $f^{-1}(S)\in\mathcal{N}$, and because $\mathcal{N}\subseteq\mathcal{M}$, we have that $f^{-1}(S)\in\mathcal{M}$.

If $0\in S$, then $F^{-1}(S)=f^{-1}(S)\cup (X\setminus U)$. Because $f$ is measurable, we have that $f^{-1}(S)\in\mathcal{N}$, hence $f^{-1}(S)\in\mathcal{M}$. Because any $\sigma$-algebra is closed under complements and unions and $U\in\mathcal{M}$, we have that $X\setminus U\in\mathcal{M}$, and hence $f^{-1}(S)\cup (X\setminus U)\in\mathcal{M}$.

Thus $F$ is measurable.

Note that if $U$ were not measurable, this argument would not work, because (as Arturo points out below) we could not conclude that $F^{-1}(S)$ would be measurable for any $S\neq\varnothing$ or $\overline{\mathbb{R}}$.

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You should probably add that $U$ should be measurable for this argument to work. –  t.b. Oct 28 '11 at 4:26
@t.b.: I thought that as well, but this follows from the premise, since $U=f^{-1}(\overline{\mathbb{R}})$ must be measurable. (On the other hand, if we are viewing $U$ as a measure space with the $\sigma$-algebra $\{U\cap M\mid M\in S\}$ (where $S$ is the $\sigma$-algebra on $X$), then I agree with you. –  Arturo Magidin Oct 28 '11 at 4:33
@Arturo: I had the second interpretation in mind. Thanks for the clarification! –  t.b. Oct 28 '11 at 4:39
Thank you so much, just right quick, what if i change $\overline{\mathbb{R}}$ to an arbitrary measurable space $Y$?? –  Mario Carrasco Oct 28 '11 at 4:46
@Mario: The exact same argument would work, with $Y$ replacing $\overline{\mathbb{R}}$ and any $y\in Y$ replacing $0$. –  Zev Chonoles Oct 28 '11 at 4:47

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