# Show a limited function is measurable

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1. Feb 11, 2017

### diddy_kaufen

Not sure about the translated term limited (from German); perhaps cut-off function?

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

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

2. Relevant equations

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

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

3. The attempt at a solution

Since $f$ is measurable, then $f_C$ is measurable when $|f(x)| < C$.

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

I'm not sure how to approach $f_C$ at $C$. Perhaps: We have shown that $f_C$ is measurable at all points except $f(x)=C$, but a single point has measure $0$. However this seems very hand-wavy and probably entirely incorrect...

2. Feb 11, 2017

### Dick

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?