There are two measurable spaces with Borel sigma-algebras on them $(X, \mathcal B (X)) $ and $(Y, \mathcal B (Y)) $. There is also a continuous function $f:(X, \mathcal B (X)) \to (Y, \mathcal B (Y)) $. Prove that this function is measurable with respect to Borel sigma-algebra.
Asked
Active
Viewed 990 times
-2
-
1fyi - this question was probably 'down-voted' (not by me!) because you didn't show any work... you might want to explain what you tried, or what is confusing you. – peter a g May 16 '16 at 16:03
-
Possible duplicate of Inverse image of Borel set – May 16 '16 at 18:20
2 Answers
2
If $(Z, \tau ) $ is topological vector space then the Borel algebra of sets is by definition the smallest $\sigma$- algebra that contains $\tau.$ So if
- A set $A\in B(Y) $ is open then by the contnuity of $f$ the set $f^{-1} (A) $ is open hence $f^{-1} (A)\in B(X).$
- Let $\mathcal{M} =\{ A\in B(Y) : f^{-1} (A) \in B(X) \}$ then $\mathcal{M} $ is sigma algebra that contains every open subset of $Y$ hence $\mathcal{M} = B(Y) .$
Thus $f$ is measurable.
0
Use definition of continuity:
$f:(X, \mathcal M) \to (Y, \mathcal N) $ is continuous iff $\ \ \forall A \subset Y \ \ A \text{ open} $ we have $f^{-1}(A)\text{ is open} $
By definition of Borel sigma algebra you get the thesis
mariob6
- 342