Consider the function [itex]g(t) = f(t)/t[/itex] on [itex][0,1][/itex], where [itex]f[/itex] is measurable on [itex][0,1][/itex]. Does it follow that [itex]g[/itex] is measurable on [itex][0,1][/itex]? I know there's a problem -- namely, division by zero -- only on a set of measure zero -- namely, [itex]\{0\}[/itex] -- and that [itex]g[/itex] agrees with the measurable function [itex]g_0 = g|_{(0,1]}[/itex] almost everywhere. So I'd think the answer is "yes," since I was under the impression that if two functions agreed almost everywhere, then if one was measurable, the other was too...but I thought that was only if the functions were defined on the same set, and [itex]g[/itex] and [itex]g_0[/itex] clearly are not.(adsbygoogle = window.adsbygoogle || []).push({});

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Measurable functions and division by zero

**Physics Forums | Science Articles, Homework Help, Discussion**