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Adventures in Measure Theory - 3

I am following the Measure Theory series by D.H. Fremlin and blogging my notes here.

Today we’re going to solve some problems!

111X (a)

The nine pairs are:

  • E∩(⋃n∈NFn)=⋃n∈N(E∩Fn)E\cap(\bigcup_{n\in\Bbb N}F_n) = \bigcup_{n\in\Bbb N}(E\cap F_n)
  • E∪(⋂n∈NFn)=⋂n∈N(E∪Fn)E\cup(\bigcap_{n\in\Bbb N}F_n) = \bigcap_{n\in\Bbb N}(E\cup F_n)
  • ⋃n∈N(En∖F)=(⋃n∈NEn)∖F\bigcup_{n\in\Bbb N}(E_n\setminus F) = (\bigcup_{n\in\Bbb N}E_n)\setminus F
  • E∖(⋃n∈NFn)=⋂n∈N(E∖Fn)E\setminus(\bigcup_{n\in\Bbb N}F_n) = \bigcap_{n\in\Bbb N}(E\setminus F_n)
  • ⋃n∈N(E∖Fn)=E∖(⋂n∈NFn)\bigcup_{n\in\Bbb N}(E\setminus F_n) = E\setminus(\bigcap_{n\in\Bbb N}F_n)
  • ⋂n∈N(En∖F)=(⋂n∈NEn)∖F\bigcap_{n\in\Bbb N}(E_n\setminus F) = (\bigcap_{n\in\Bbb N}E_n)\setminus F
  • ⋂m,n∈N(Em∖Fn)=(⋂n∈NEn)∖(⋃n∈NFn)\bigcap_{m,n\in\Bbb N}(E_m\setminus F_n) = (\bigcap_{n\in\Bbb N}E_n)\setminus(\bigcup_{n\in\Bbb N}F_n)
  • (⋃n∈NEn)∩(⋃n∈NFn)=⋃m,n∈N(Em∩Fn)(\bigcup_{n\in\Bbb N}E_n)\cap(\bigcup_{n\in\Bbb N}F_n) = \bigcup_{m,n\in\Bbb N}(E_m\cap F_n)
  • (⋂n∈NEn)∪(⋂n∈NFn)=⋂m,n∈N(Em∪Fn)(\bigcap_{n\in\Bbb N}E_n)\cup(\bigcap_{n\in\Bbb N}F_n) = \bigcap_{m,n\in\Bbb N}(E_m\cup F_n)

111X (b)

All ‘open intervals’ (a,b)(a,b), (−∞,b)(-\infty, b), (a,∞)(a, \infty) are open sets.

Proof.
Consider an interval (a,b)(a,b). We want to show that ∀  x∈(a,b)  ∃  δ>0\forall\,\, x\in (a,b)\,\,\exists\,\,\delta > 0 such that (x−δ,x+δ)∈(a,b)(x-\delta,x+\delta)\in (a,b). If we set δ=min⁡{x−a,b−x}\delta = \min\{x-a, b-x\} we are done.

  • How did we get this value of δ\delta?
    Since δ\delta will be different for every xx, we can deduce that it must be a function of xx. Now we want a value of δ\delta such that x−δ≥ax - \delta \ge a and x+δ≤bx + \delta \le b. The largest permissible value of δ\delta will be the minimum of the two values we get when these two inequalities are written as equalities i.e. either x−δ=ax - \delta = a or x+δ=bx + \delta = b.
All intervals (bounded or unbounded, open, closed or half-open) are Borel sets.

Proof.
Borel sets are the members of the σ\sigma-algebra of subsets of R\Bbb R generated by the open sets of R\Bbb R. We just proved above that all open intervals (bounded and unbounded) are open sets, and hence are also Borel sets. Now we have to prove that closed and half-open intervals are also Borel sets.

For any open interval (a,b)(a, b) in a Borel σ\sigma-algebra, it’s complement (−∞,a]∪[b,∞)(-\infty,a]\cup[b, \infty) also lies in the Borel σ\sigma-algebra. Likewise we can construct all possible half-open or closed intervals simply by taking the union, complement or perhaps both of the Borel sets we already have (or get).

There must be a more elegant proof; I am not convinced with the two paragraphs I have whipped up. Based on what I found here on StackExchange, here is a much better proof:

The initial argument remains the same. All open intervals must be Borel sets (by definition). We will now show that closed and half-open intervals can be expressed as open intervals.

Consider the sets [a,b][a, b] and ⋂n∈N(a−1n,b+1n)\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n}). It is clear that [a,b]⊆(a−1n,b+1n)∀n∈N[a, b] \subseteq (a-\frac{1}{n}, b+\frac{1}{n})\forall n\in\Bbb N. Thus,

[a,b]⊆⋂n∈N(a−1n,b+1n)(i)[a, b] \subseteq\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n}) \tag{i}

Now, let x∈⋂n∈N(a−1n,b+1n)x\in\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n}). We will prove that x∈[a,b]x\in [a, b] by contradiction. Let us assume that x∉[a,b]x\notin [a, b] or more specifically, x<ax < a. Then, by the Archimedean property, we can say that ∃  n∈N\exists\,\, n\in\Bbb N such that 1<n(a−x)1 < n(a-x) or x<a−1nx < a - \frac{1}{n}. Similarly, if x>bx > b then x>b+1nx > b+\frac{1}{n} for some n∈Nn\in\Bbb N. These two inequalities are a contradiction because a−1n<x<b+1n  ∀  n∈Na-\frac{1}{n} < x < b+\frac{1}{n} \,\,\forall\,\, n\in\Bbb N. This means that our assumption was wrong and x∈[a,b]x\in [a, b] indeed. Thus,

⋂n∈N(a−1n,b+1n)⊆[a,b](ii)\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n})\subseteq [a, b]\tag{ii}

From (i)(i) and (ii)(ii), we have,

[a,b]=⋂n∈N(a−1n,b+1n)[a, b] = \bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n})

This means that [a,b][a, b] is also an open interval and hence a Borel set.

In a similar fashion we can prove that,

(a,b]=⋂n∈N(a,b+1n)[a,b)=⋂n∈N(a−1n,b)(a, b] = \bigcap_{n\in\Bbb N} (a, b + \frac{1}{n}) \\ [a, b) = \bigcap_{n\in\Bbb N} (a - \frac{1}{n}, b)

Hence, proved.