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Measurable functions and division by zero

  1. Aug 17, 2010 #1
    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.
  2. jcsd
  3. Aug 17, 2010 #2


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    There is nothing in the definition of measurable functions to prevent a function from being infinite.
  4. Aug 19, 2010 #3


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    In other words, g is defined on the whole interval [0,1]:

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