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

Problem 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:

$$f_C(x) = \left\{ \begin{array}{ll} f(x) & \mbox{if } |f(x)| \leq C \\ C & \mbox{if } f(x) > C \\ -C & \mbox{if } f(x) < -C \end{array} \right.$$

Clarification

By measure space it is meant: 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.

My attempt

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.

1 Answers 1

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I'm sure you have a theorem that say that a product and a sum of meaurable function is measurable. Remark that

$$f_C(x)=C\boldsymbol 1_{f(x)>C}(x)-C\boldsymbol 1_{f(x)<-C}(x)+f(x)\boldsymbol 1_{f(x)\in [-C,C]}(x),$$ allow you to conclude very easily.