Michelle Mendoza

Answered

2022-07-01

Let $$ be some measurable domain and $f:E\to \mathbb{R}$ a measurable map. Let $B\in \mathrm{Bor}(\mathbb{R})$ be a Borel set. Show that ${f}^{-1}(B)$ is measurable.

I'm advised to define $\mathcal{A}=\{A\mid {f}^{-1}(A)\text{measurable}\}$. Now $\mathcal{A}$ consists of sets whose preimage is measurable, and since $f$ is continuous these sets are open. This collection seems to form a $\sigma $-algebra on $\mathbb{R}$, but I'm confused about the construction here as it seems that $\mathcal{A}$ is the smallest $\sigma $-algebra containing open sets, but that would mean that it's equal to $\mathrm{Bor}(\mathbb{R})$?

I'm advised to define $\mathcal{A}=\{A\mid {f}^{-1}(A)\text{measurable}\}$. Now $\mathcal{A}$ consists of sets whose preimage is measurable, and since $f$ is continuous these sets are open. This collection seems to form a $\sigma $-algebra on $\mathbb{R}$, but I'm confused about the construction here as it seems that $\mathcal{A}$ is the smallest $\sigma $-algebra containing open sets, but that would mean that it's equal to $\mathrm{Bor}(\mathbb{R})$?

Answer & Explanation

Freddy Doyle

Expert

2022-07-02Added 20 answers

Your proof will depend on your definition of "measurable function".

Let's say "measurable function" means ${f}^{-1}(G)$ is measurable for all open sets $G\subseteq \mathbb{R}$.

Then define $\mathcal{A}=\{A\subseteq \mathbb{R}\mid {f}^{-1}(A)\text{measurable}\}$. Show that $\mathcal{A}$ contains all open sets and that $\mathcal{A}$ is a sigma-algebra. Conclude that $\mathcal{A}\supseteq \mathrm{Bor}(\mathbb{R})$.

Let's say "measurable function" means ${f}^{-1}(G)$ is measurable for all open sets $G\subseteq \mathbb{R}$.

Then define $\mathcal{A}=\{A\subseteq \mathbb{R}\mid {f}^{-1}(A)\text{measurable}\}$. Show that $\mathcal{A}$ contains all open sets and that $\mathcal{A}$ is a sigma-algebra. Conclude that $\mathcal{A}\supseteq \mathrm{Bor}(\mathbb{R})$.

Pattab

Expert

2022-07-03Added 2 answers

Note: you cannot prove $\mathcal{A}=\mathrm{Bor}(\mathbb{R})$.

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