1

1.1 (a)

\(f:\mathbb{R}\to\mathbb{R}\) is continuous, for any open subset \(U\subseteq \mathbb{R}\), \(f^{-1}(U)\) is open and therefore Borel.

1.2 (b)

\(g:\mathbb{R}\to\mathbb{R}\) is continuous everywhere except at \(0\), for any open \(U\subseteq \mathbb{R}\), let \(V = U\setminus\left\{g(0)\right\}\) which is open. Suppose \(U = V\sqcup\left\{g(0)\right\}\) (other case \(U=V\) is trivial), \[g^{-1}(U) = g^{-1}(V)\cup g^{-1}(\left\{g(0)\right\}),\] \(g^{-1}(V)\) is open because \(g\) is continuous on \(V\), while \(g^{-1}(\left\{g(0)\right\}\) is the union of \(\left\{0\right\}\) and a closed set.

2

Basically use definition of continuous (pullback of open is open), so taking \(f^{-1}\) of a Borel set keeps it Borel.

3

It’s 1.

4

1. Any finite covering of \(E\) with closed intervals also covers \(\overline{E}\).

2. \([0,1]\cap\mathbb{Q}\) should do the trick. If I use finitely many intervals their length must add up to 1.

5

Let \(\epsilon>0\), then do analysis.