2
$\begingroup$

Let $f,g$ be $\mathcal E$-$\mathcal B(\mathbb R)$-measurable functions. I want to show piecewise function $h$ of $f$ and $g$ is also measurable.

Suppose $(X, \mathcal E)$ is a measure space, let $f,g$ be $\mathcal E$-$\mathcal B(\mathbb R)$-measurable functions and let $A \in \mathcal E$.

I want to show $h: X \rightarrow \mathbb R$ given by $ h(x) = \left\{ \begin{array}{lr} f(x) : x \in A\\ g(x) : x \in A^C \end{array} \right.\\$ is again a $\mathcal E$-$\mathcal B(\mathbb R)$-measurable function.

I've tried writing $(-\infty, a]$ as two disjoint sets $A_1, A_2$ such that $A_1 \cup A_2 = (-\infty, a]$, but then $f^{-1}(-\infty, a]) = f^{-1}(A_1 \cup A_2) = f^{-1}(A_1) \cup f^{-1}(A_2)$ and I can't say whether this is an element of $\mathcal E$. Also I don't use that $A \in \mathcal E$.

Can anyone help ?

$\endgroup$
  • $\begingroup$ Can I just check your notation; you mean f and g are measurable functions from (X, E) to (R,Borel)? $\endgroup$ – Harry Wilson Sep 12 '14 at 11:58
  • $\begingroup$ Yes, exactly :) $\endgroup$ – Shuzheng Sep 12 '14 at 12:05
3
$\begingroup$

For any Borel set $B$ one has $$ h^{-1}(B)=\big(f^{-1}(B)\cap A\big)\cup \big(g^{-1}(B)\cap A^c\big). $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ahh, and now the result follow easily by means of every intersection and union lies in $\mathcal E$. Thanks ! $\endgroup$ – Shuzheng Sep 12 '14 at 12:02
  • $\begingroup$ Yep. You're welcome. $\endgroup$ – Stefan Hansen Sep 12 '14 at 12:03
2
$\begingroup$

Do you know that multiplication and addition of measurable functions are again measurable? If yes, simply note that

$$ h = f \cdot \chi_A + g \cdot \chi_{A^c}, $$

where $\chi_A$ is the characteristic function of $A$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.