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Show a limited function is measurable

  1. Feb 11, 2017 #1
    Not sure about the translated term limited (from German); perhaps cut-off function?

    1. The problem statement, all variables and given/known data

    Let [itex]f[/itex] be a measurable function in a measure space [itex](\Omega, \mathcal{F}, \mu)[/itex] and [itex]C>0[/itex]. Show that the following function is measurable:
    [tex]f_C(x) =
    \left\{
    \begin{array}{ll}
    f(x) & \mbox{if } |f(x)| \leq C \\
    C & \mbox{if } f(x) > C \\
    -C & \mbox{if } f(x) < -C
    \end{array}
    \right.[/tex]

    2. Relevant equations

    None in particular. Definition, measurable space:
    An ordered tuple [itex](\Omega,\mathcal{F})[/itex], where [itex]\Omega[/itex] is a set and [itex]\mathcal{F}[/itex] is a [itex]\sigma[/itex]-algebra of subsets in [itex]\Omega[/itex], is called a *measurable space*.

    Definition, measureable function:
    https://en.wikipedia.org/wiki/Measurable_function

    3. The attempt at a solution

    Since [itex]f[/itex] is measurable, then [itex]f_C[/itex] is measurable when [itex]|f(x)| < C[/itex].

    It should be trivial to prove that a constant function is measurable.

    I'm not sure how to approach [itex]f_C[/itex] at [itex]C[/itex]. Perhaps: We have shown that [itex]f_C[/itex] is measurable at all points except [itex]f(x)=C[/itex], but a single point has measure [itex]0[/itex]. However this seems very hand-wavy and probably entirely incorrect...
     
  2. jcsd
  3. Feb 11, 2017 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can make it a lot less hand-wavy if you state and work with the definition of 'measurable'. You want to show ##\{x:f_C(x)<a\}## is measurable for all ##a##. You know ##\{x:f(x)<a\}## is measurable for all ##a##. How do they compare?

     
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