# Thread: Measurable function in intersection set

1. ## Measurable function in intersection set

If $X = A \cup B$ where $A,B \in \mathbb {M}$, the $\sigma$-Algebra of X, then a function f on X is measurable iff f is measurable on A and on B.

Proof so far.

Assume that $f:X \rightarrow Y$, where $\mathbb {M} , \mathbb {N}$ are the set of all $\sigma$-Algebra of X and Y, respectively.

First, assume that f is measurable on X, that is, I have $f^{-1}(E) \in \mathbb {M} \ \ \ \ \forall E \in \mathbb {N}$

Consider the function $f:A \rightarrow Y$, I need to show that $f^{-1}(E) \in \mathbb {M}_A \ \ \ \forall E \in \mathbb {N}$, where $\mathbb {M}_A$ is the a $\sigma$-Algebra of A.

I know that has to be true, but I'm just having a hard time trying to write up the sigma algebra of A in X, any hints?

Thank you.

If $X = A \cup B$ where $A,B \in \mathbb {M}$, the $\sigma$-Algebra of X, then a function f on X is measurable iff f is measurable on A and on B.

Proof so far.

Assume that $f:X \rightarrow Y$, where $\mathbb {M} , \mathbb {N}$ are the set of all $\sigma$-Algebra of X and Y, respectively.

First, assume that f is measurable on X, that is, I have $f^{-1}(E) \in \mathbb {M} \ \ \ \ \forall E \in \mathbb {N}$

Consider the function $f:A \rightarrow Y$, I need to show that $f^{-1}(E) \in \mathbb {M}_A \ \ \ \forall E \in \mathbb {N}$, where $\mathbb {M}_A$ is the a $\sigma$-Algebra of A.

I know that has to be true, but I'm just having a hard time trying to write up the sigma algebra of A in X, any hints?
If f is measurable on A then $f^{-1}(E)\cap A \in \mathbb{M}$. If f is measurable on B then $f^{-1}(E)\cap B \in \mathbb{M}$. It then follows that $f^{-1}(E) = f^{-1}(E)\cap (A\cup B) = (f^{-1}(E)\cap A)\cup(f^{-1}(E)\cap B) \in \mathbb{M}$.