4. 99% lemma (-ish)

problem

Prove that if $μ$ is a finite measure on $𝒞$ and $A⊆𝒞$ is a $μ$-measurable set such that $μ(A)>0$, then for every $ϵ>0$, there is a non-null finitely-determined set $U⊆𝒞$ with

$$\frac {μ(A \mathbin\triangle U)} {μ(U)} ≤ ϵ.$$

proof

definition

For each finite string $p∈\{0,1\}^*$, let $U_p$ denote the set of strings with prefix $p$. Recall that $\{Uₚ\}_{p∈\{0,1\}^*}$ forms a basis for the topology of $𝒞$.

Fix $ϵ>0$. Let $δ>0$ be small enough so that $δ(2+δ)≤ϵ$. By regularity of measure, there exists an open set $V$ so that $A⊆V$ and

$$μ(V)≤(1+δ)\,μ(A).$$

Write $V$ as a union of the basis elements; that is, let $P⊆\{0,1\}^*$ so that $V=⋃_{p∈P}Uₚ$. Assume without loss of generality that there do not exist $p,q∈P$ with $p$ a proper prefix of $q$.

aside

To see why we can assume this, let $\tilde P=\{q∈P\mid∄p:\text{\(p\) is proper prefix of \(q\)}\}$ and observe that for every $q∈P∖\tilde P$, since there exists $p$ a proper prefix of $q$ we already have $Uₚ⊃U_q$ in the union, so $⋃_{p∈\tilde P}Uₚ=⋃_{p∈P}Uₚ$.

Then notice that for all $p,q∈P$, since neither is a proper prefix of the other they must differ at a common index, so $U_p∩U_q=∅$. Therefore $μ(V)=∑_{p∈P}μ(Uₚ)$. This is either just a finite sum (if $P$ itself is finite) or a sum converging to a finite positive value; either way, there exists a finite $P'⊆P$ so that letting $U=⋃_{p∈P'}Uₚ$ we have

$$μ(U) ≥ (1-δ)\,μ(V).$$

Note that $U⊆V$ and also that $U$ is finitely-determined.

Finally, we have

$$A\mathbin\triangle U = \underbrace{(A∖U)}_{⊆V∖U} ∪ \underbrace{(U∖A)}_{⊆V∖A},$$

and so

$$μ(A\mathbin\triangle U) ≤ [μ(V)-μ(U)] + [μ(V)-μ(A)].$$

Recall that $V$ and $U$ were constructed to satisfy

$$\frac {μ(V)} {μ(A)} ≤ 1+δ, \qquad \frac {μ(U)} {μ(V)} ≥ 1-δ,$$

so $μ(V)/μ(U)≤1/(1-δ)≤1+δ$, and

$$\begin{aligned} \frac {μ(A\mathbin\triangle U)} {μ(U)} &≤ \underbrace{\left(\frac {μ(V)} {μ(U)} - 1\right)}_{≤δ} + \underbrace{\left(\frac {μ(V)} {μ(A)} - 1\right)}_{≤δ} \frac {μ(A)} {μ(U)} \\ &≤ δ \left(1 + \frac {μ(A)} {μ(U)}\right) \\ &≤ δ \left(1 + \frac {μ(V)} {μ(U)}\right) \\ &≤ δ (2 + δ) ≤ ϵ. \\ \end{aligned}$$