- #1

diddy_kaufen

- 7

- 0

*limited*(from German); perhaps

*cut-off*function?

## Homework Statement

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]

## Homework 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

## 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...