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) &amp; \mbox{if } |f(x)| \leq C \\<br /> C &amp; \mbox{if } f(x) &gt; C \\<br /> -C &amp; \mbox{if } f(x) &lt; -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)| &lt; 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&gt;0. Show that the following function is measurable:
f_C(x) =<br /> \left\{<br /> \begin{array}{ll}<br /> f(x) &amp; \mbox{if } |f(x)| \leq C \\<br /> C &amp; \mbox{if } f(x) &gt; C \\<br /> -C &amp; \mbox{if } f(x) &lt; -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)| &lt; 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...