Measurable functions and division by zero

[AxiomOfChoice]

Consider the function g(t) = f(t)/t on [0,1], where f is measurable on [0,1]. Does it follow that g is measurable on [0,1]? I know there's a problem -- namely, division by zero -- only on a set of measure zero -- namely, \{0\} -- and that g agrees with the measurable function g_0 = g|_{(0,1]} 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 g and g_0 clearly are not.

 

[mathman]

There is nothing in the definition of measurable functions to prevent a function from being infinite.

 

[Landau]

In other words, g is defined on the whole interval [0,1]:
g:[0,1]\to\overline{\mathbb{R}}

taking the value \infty at t=0.
(well, I don't know how f is defined, but at least if f if is everywhere finite this makes sense)