Borel measurable function $f : X \rightarrow \mathbb{C}$.

Question

In some old books I found the following definition:

"Given $f : X \rightarrow \mathbb{C}$, $f$ is a Borel measurable function if $f^{-1}(B)$ is a Borel set [of X] for any open set $B \subseteq \mathbb{C}$."

My question is: if we consider the Borel $\sigma-$algebra for $X$ and also for $\mathbb{C}$, is enough to check the condition of the definition to show that a function is measurable with respect to these two $\sigma-$algebras?

I would be truly grateful if you could give some references.

Thank you

Welcome to MSE. A hint: the preimage preserves unions, intersections and complements.See Overview of basic results about images and preimages. — user539887, May 09 '18 at 10:55
Thank you, I was thinking about that, but I was not pretty sure. — Glazunov, May 09 '18 at 20:01

score 2 · Accepted Answer · answered May 09 '18 at 10:51

2

Yes. Let $X$, $Y$ non-empty sets, $A$ a $ \sigma-$ algebra on $X$, $B$ a $ \sigma-$ algebra on $Y$ and $f:X \to Y$ a mapping. If $C$ is a collection of subsets of $B$ such that $\sigma(C)=B$, then we have:

$f$ is $(A,B)-$ measurable $ \iff f^{-1}(M) \in A$ for all $M \in C$.

answered May 09 '18 at 10:51

Fred

77,394

Thank you very much. I have just found this result in "Measure Theory" by Donald Cohn :) – Glazunov May 09 '18 at 10:55
1

..... a very fine book .... – Fred May 09 '18 at 10:57

Borel measurable function $f : X \rightarrow \mathbb{C}$.

1 Answers1