Truncation of a function is measurable

Question

Let $f$ be a measurable function, then the truncation of $f$

$$ f^a(x) = \begin{cases} a & \text{if }f(x) > a \\ f(x) & \text{if }f(x) \leq a \end{cases} $$

is a measurable function.

Im trying to solve this problem, but I am having some dificulties. For instance, I want to show $(f^a)^{-1} ((b, \infty)) $ in measurable. If $ b < a$, then $(f^a)^{-1} ((b, \infty)) = \mathbb{R}$ which is measurable. But if $b \geq a$, then $(f^a)^{-1} ((b, \infty)) = f^{-1}(-\infty, a) \cup f^{-1}(a, \infty). $ Is this a correct solution? Im kind of unsecure about last part. Any help would greatly be welcome.

$f^a = a\chi_{(a,\infty)} + f\chi_{(-\infty,a]}$ – Prahlad Vaidyanathan Oct 06 '13 at 05:08 — Prahlad Vaidyanathan, Oct 06 '13 at 05:08

score 2 · Answer 1 · answered Jun 21 '14 at 01:42

2

Write the truncation as $\min(f,a)$ and use the fact that if two functions are measurable, so is their minimum...

answered Jun 21 '14 at 01:42

user127474

421

Truncation of a function is measurable

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