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Spring 2017 - Problem 3

Banach spaces, topology

Let C([0,1])C\p{\br{0,1}} denote the Banach space of continuous functions on the interval [0,1]\br{0, 1} endowed with the sup-norm. Let F\mathcal{F} be a σ\sigma-algebra on C([0,1])C\p{\br{0,1}} so that for all x∈[0,1]x \in \br{0, 1}, the map defined via

Lx(f)=f(x)L_x\p{f} = f\p{x}

is F\mathcal{F}-measurable. Show that F\mathcal{F} contains all open sets.
Solution.

Fix f∈C([0,1])f \in C\p{\br{0,1}} and let B={g∈C([0,1])∣∥f−g∥≥r}B = \set{g \in C\p{\br{0,1}} \mid \norm{f - g} \geq r} for some r>0r > 0. Observe that by definition, ∥f−g∥≥r\norm{f - g} \geq r occurs if and only if for any ε>0\epsilon > 0, there exists x∈[0,1]x \in \br{0, 1} such that ∣f(x)−g(x)∣≥r−ε\abs{f\p{x} - g\p{x}} \geq r - \epsilon. We parametrize ε\epsilon via 1n\frac{1}{n} and by appealing to density of Q\Q and continuity of f−gf - g, we may replace x∈[0,1]x \in \br{0, 1} with q∈[0,1]∩Qq \in \br{0, 1} \cap \Q in the statement. Notice then that

∣f(q)−g(q)∣≥r−1n  ⟺  g(q)−f(q)≥r−1n or g(q)−f(q)≤−r+1n  ⟺  g(q)≥f(q)+r−1n or g(q)≤f(q)−r+1n  ⟺  Lq(g)≥Lq(f)+r−1n or Lq(g)≤Lq(f)−r+1n  ⟺  g∈Lq−1([Lq(f)+r−1n))∪Lq−1((−∞,Lq(f)−r+1n]).\begin{aligned} \abs{f\p{q} - g\p{q}} \geq r - \frac{1}{n} &\iff g\p{q} - f\p{q} \geq r - \frac{1}{n} \text{ or } g\p{q} - f\p{q} \leq -r + \frac{1}{n} \\ &\iff g\p{q} \geq f\p{q} + r - \frac{1}{n} \text{ or } g\p{q} \leq f\p{q} - r + \frac{1}{n} \\ &\iff L_q\p{g} \geq L_q\p{f} + r - \frac{1}{n} \text{ or } L_q\p{g} \leq L_q\p{f} - r + \frac{1}{n} \\ &\iff g \in L_q^{-1}\p{\pco{L_q\p{f} + r - \frac{1}{n}}} \cup L_q^{-1}\p{\poc{-\infty, L_q\p{f} - r + \frac{1}{n}}}. \end{aligned}

Hence,

g∈B  ⟺  ∥f−g∥≥r  ⟺  ∀n≥1,∃q∈[0,1]∩Q:∣f(q)−g(q)∣≥r−1n  ⟺  ∀n≥1,∃q∈[0,1]∩Q:g∈Lq−1([Lq(f)+r−1n))∪Lq−1((−∞,Lq(f)−r+1n])  ⟺  g∈⋂n=1∞⋃q∈[0,1]∩Q(Lq−1([Lq(f)+r−1n))⏟∈F∪Lq−1((−∞,Lq(f)−r+1n])⏟∈F).\begin{aligned} g \in B &\iff \norm{f - g} \geq r \\ &\iff \forall n \geq 1, \exists q \in \br{0, 1} \cap \Q : \abs{f\p{q} - g\p{q}} \geq r - \frac{1}{n} \\ &\iff \forall n \geq 1, \exists q \in \br{0, 1} \cap \Q : g \in L_q^{-1}\p{\pco{L_q\p{f} + r - \frac{1}{n}}} \cup L_q^{-1}\p{\poc{-\infty, L_q\p{f} - r + \frac{1}{n}}} \\ &\iff g \in \bigcap_{n=1}^\infty \bigcup_{q \in \br{0,1}\cap\Q} \p{\underbrace{L_q^{-1}\p{\pco{L_q\p{f} + r - \frac{1}{n}}}}_{\in\mathcal{F}} \cup \underbrace{L_q^{-1}\p{\poc{-\infty, L_q\p{f} - r + \frac{1}{n}}}}_{\in\mathcal{F}}}. \end{aligned}

Notice that the set in the right-hand side is comprised of countable unions and intersections, hence in F\mathcal{F}, so B∈FB \in \mathcal{F}. Since F\mathcal{F} is closed under complements, it follows that open balls are contained in F\mathcal{F}.

To complete the proof, it suffices to show that open balls generate the topology on C([0,1])C\p{\br{0, 1}}. By Stone-Weierstrass, C([0,1])C\p{\br{0, 1}} is separable (e.g., via polynomials on [0,1]\br{0, 1}), so let {fn}n\set{f_n}_n be a countable dense subset of F\mathcal{F}. Let

U={B(fn,q)∣n≥1, q∈Q∩(0,∞)}⊆F,\mathcal{U} = \set{B\p{f_n, q} \mid n \geq 1,\ q \in \Q \cap \p{0, \infty}} \subseteq \mathcal{F},

which is countable. Now let U⊆C([0,1])U \subseteq C\p{\br{0, 1}} be an arbitrary open set. Then for any f∈Uf \in U, there exists δ>0\delta > 0 such that B(f,δ)⊆UB\p{f, \delta} \subseteq U. By density, there exists a rational q<δ4q < \frac{\delta}{4} and n≥1n \geq 1 such that ∥f−fn∥<q\norm{f - f_n} < q and so given any g∈B(fn,q)∈Ug \in B\p{f_n, q} \in \mathcal{U}, we have

∥f−g∥≤∥f−fn∥+∥fn−g∥≤2q<δ2<δ,\norm{f - g} \leq \norm{f - f_n} + \norm{f_n - g} \leq 2q < \frac{\delta}{2} < \delta,

so B(fn,q)⊆UB\p{f_n, q} \subseteq U. Thus, we may write UU as a union of open balls in U\mathcal{U}, hence as a countable union of open balls. Since F\mathcal{F} is a σ\sigma-algebra, it follows that U∈FU \in \mathcal{F}, which completes the proof.