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)