Show a limited function is measurable

[diddy_kaufen]

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

Homework Statement

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

Homework 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

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...

 

[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?

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

Homework Statement

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

Homework 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

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...