The measurability of truncation function

Question

Show directly that if $f$ is measurable and $A>0$, then the truncation $f_A$ defined by $$f_A(x)=\begin{cases}f(x)&\text{ if }|f(x)| \le A,\\ A &\text{ if }f(x) > A,\\-A&\text{ if }f(x) < -A\end{cases}$$ is measurable.

"$f$ is measurable" means that $\{x \in X | f(x) > \alpha\} \in \sum$. If $f$ is Borel measurable, I can divide the interval into $(-\infty, -A) \cup[-A,A] \cup (A, \infty).$ By doing this, I can prove that $f_A$ is measurable since the interval of $f_A$ is the complement of $f$ in $(-\infty, -A)\cup(A, \infty)$ . However, if I don't know the interval of $\sum$, how can I approach this question? Could you give some help?

Thank you in advance.

Answer 1

We have $f_A(x)=f(x) \cdot 1_{\{|f(x)| \le A\}}+A \cdot 1_{\{f(x)>A\}}-A\cdot 1_{\{f(x)>-A\}}$.

Each function on the right is measurable ($1_B$ denotes the char. function of the set $B$).

Answer 2

Let

$$\phi (x)=\begin{cases}x,&\text{if }|x| \leq A,\\A,&\text{if }x>A,\\-A,&\text{it }x<-A, \end{cases}.$$

Then $f_A=\phi \circ f$ which is a composition of two measurable functions, hence measurable. [ $\phi$ is continuous, hence Borel measurable].