| title | author | date |
|---|---|---|
MA4262 Tutorial 2 |
Qi Ji |
8th September 2020 (LATE) |
\newcommand{\set}[1]{\left{#1\right}} \newcommand{\B}{\mathcal{B}} \newcommand{\R}{\mathbb{R}}
(f:\R\to\R) is continuous, for any open subset (U\subseteq \R), (f^{-1}(U)) is open and therefore Borel.
(g:\R\to\R) is continuous everywhere except at (0), for any open (U\subseteq \R), let (V = U\setminus\set{g(0)}) which is open. Suppose (U = V\sqcup\set{g(0)}) (other case (U=V) is trivial), [g^{-1}(U) = g^{-1}(V)\cup g^{-1}(\set{g(0)}),] (g^{-1}(V)) is open because (g) is continuous on (V), while (g^{-1}(\set{g(0)}) is the union of (\set{0}) and a closed set.
Basically use definition of continuous (pullback of open is open), so taking (f^{-1}) of a Borel set keeps it Borel.
It's 1.
a. Any finite covering of (E) with closed intervals also covers (\overline{E}).
b. ([0,1]\cap\mathbb{Q}) should do the trick. If I use finitely many intervals their length must add up to 1.
Let (\epsilon>0), then do analysis.