# 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.

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 ?

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

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

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

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 $A$.