# Extension of measurable function defined on dense subset.

I'm looking for some books or papers written about the problem below but I don't find them.

Problem

Let $(a,b)\subset\mathbb{R}$ be an open interval and $I$ be a subset of $(a,b)$ such that $I$ is dense in $(a,b)$. Also let $f:I\to\mathbb{R}$ be a (Lebesgue) measurable function. Then, does a (Lebesgue) measurable function $g:(a,b)\to\mathbb{R}$ exist uniquely so that $g|_{I}=f$?

How is such a problem know? I wonder if you tell me references or informations you know.

• An interval $I$ can't be dense in $(a,b)$ unless it's equal to $(a,b)$. Perhaps you mean $f$ to be defined on a measurable subset of $(a,b)$? But the extension is far from unique: e.g. you could take $g$ to be any constant on the complement of that set. – Robert Israel Apr 6 '16 at 6:37
• @RobertIsrael Thank you for pointing out. That's a typo so I'll edit. By the way, what is a sufficient condition to guarantee a uniqueness? Is it that $I$ has a full measure $(b-a)$ for instance? – user Apr 6 '16 at 6:40
• You can always change $g$ at a single point, for example, so you never have uniqueness unless $I = (a,b)$. – Robert Israel Apr 6 '16 at 15:25