1

Suppose \(E\in\mathcal{X}\), pullback an open set \(U\subseteq \mathbb{R}\), 4 cases

• if \(U\) contains neither \(0\) nor \(1\) preimage is empty (and measureable),
• similar story for containing both, preimage will be \(X\),
• if \(U\) contains only \(0\), the preimage is \(E^c\),
• likewise if \(U\) contains only \(1\) the preimage is \(E\).

If \(E\notin\mathcal{X}\) we take pre-image of \(\left\{1\right\}\) which is just \(E\) which by hypothesis is not \(\mathcal{X}\)-measurable.

2

Idea: iterate through all (finitely many) nonempty subsets \(S \subseteq \left\{E_1,\dots,E_k\right\}\), define \(F_S\) to be the intersection and let \(c_s = \sum_{E_k\in S} b_k\).

Show that the two functions coincide, show that disjoint refinement is unique-ish (so if two integrals differ the functions differ), disjoint refinement and additivity of measure helps here.

3

Finite times finite is finite.

4

1. Easy to see mapping is one-one. Also easy to see that \(c\varphi(x) \leq cf(x)\).

2. Sup both sides.

5

damn technical

6

no time sry