1

Let $(X,\mathcal M)$ be a measurable space. Are these two equivalent?

  1. $f:(X,\mathcal M)\to (\mathbb R,\mathcal B(\mathbb R))$ is measurable.

  2. $f:(X,\mathcal M)\to (\mathbb R,\{(a,\infty):a\in\mathbb R\})$ is measurable.

Here, $\mathcal B(\mathbb R)$ is the collection of all borel sets in $\mathbb R$. To say that $f:(X,\mathcal M)\to (Y,\mathcal N)$ is measurable means $f:X\to Y$ and $f^{-1}(B)\in\mathcal M$ for all $B\in\mathcal N$. (I know that $\{(a,\infty):a\in\mathbb R\}$ is not a $\sigma$-algebra, but let's not worry about that.)

I'm curious about this because in the book Rudin. "Principles of Mathematical Analysis" 3/e. p. 310. (Chapter 11. The Lebesgue Theory), the author defines a measurable function to be a function that satisfies 2, while other books use 1 to define a measurable function.

It's easy to prove that 1 implies 2, since $(a,\infty)\in\mathcal B(\mathbb R)$ for all $a\in\mathbb R$. But I want to know if 2 implies 1.

zxcv
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  • 2 implies 1 because the smallest $\sigma$-algebra containing the sets $(a,\infty)$ contains the standard topology on $\mathbb R$, so the two generated $\sigma$-algebras are the same. – Ashwin Trisal Aug 27 '18 at 06:54

2 Answers2

1

2 implies 1.

Let it be that for every $a\in\mathbb R$ we have $f^{-1}((a,\infty))\in\mathcal M$ or equivalently that $f^{-1}(\mathcal V)\subseteq\mathcal M$ where $\mathcal V:=\{(a,\infty)\mid a\in\mathbb R\}$.

Then: $$\sigma(f^{-1}(\mathcal V))\subseteq\mathcal M\tag1$$ and it can be shown that in general:$$\sigma(f^{-1}(\mathcal V))=f^{-1}(\sigma(\mathcal V))\tag2$$

For a proof of that see here.

So based on $(1)$ and $(2)$ we find:$$f^{-1}(\sigma(\mathcal V))\subseteq\mathcal M$$

The final step is proving that:$$\sigma(\mathcal V)=\mathcal B(\mathbb R)$$

It is evident that $\sigma(\mathcal V)\subseteq\mathcal B(\mathbb R)$ and for the other side it is enough to prove that:$$\tau\subseteq\sigma(\mathcal V)$$where $\tau$ denotes the standard topology on $\mathbb R$.

For this realize that every open set in $\mathbb R$ can be written as countable union of open intervals, secondly that $(a,b]=(a,\infty)\cap(b,\infty)^{\complement}$ and thirdly that $(a,b)=\bigcup_{q\in\mathbb Q\cap(a,b)}(a,q]$.

drhab
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0

(2)$\implies $(1) is obvious. For the converse, consider $$\mathcal A=\{X\subset \mathbb R\mid f ^{-1}(X)\in \mathcal M\}.$$

It's a $\sigma -$algebra that contain open set (why ?) and thus $\mathcal B(\mathbb R)\subset \mathcal A$ what prove the claim.

Surb
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