Math 597. Homework 6 Bryden Cais

(1) Let f : E → R^{∗} be measurable and assume that |f| > 0 a.e. on E. Then for any ϵ > 0 there exists

a measurable set E_{ϵ} ⊂ E and a positive number δ_{ϵ} such that μ(E \ E_{ϵ}) ≤ ϵ and |f| > δ_{ϵ} on E_{ϵ}.

Without loss of generality, we may suppose that f is finite a.e. since if it is not, then we already have |f| > δ_{ϵ} on some X ⊂ E where f is not finite on X, so we need only be concerned with the subset of E on which f is a.e. finite.

Now define

g_{n}(x) = \begin{cases}\frac{1}{f(x)} & if |f(x)| ≥ 1/n \\ 1 & otherwise \end{cases} .

Clearly, g_{n} → 1/f pointwise a.e. and since |f| > 0 a.e, we have 1/f finite a.e. Since E is of finite measure, for any ϵ > 0, the Egorov Theorem guarantees the existence of a set E_{ϵ} ⊂ E on which the convergence g_{n} → 1/f is uniform. That is, for any δ > 0 there exists an N > 0 such that for all n ≥ N, |g_{n}(x) − 1/f(x)| < δ for all x ∈ E_{ϵ}. Now by construction, we have |g_{n}| < n so that for any δ > 0 there exists an N such that for all n ≥ N we have |g_{n} − f| < δ, so that |f| < |g_{n}| + δ ≤ N + δ. Thus, for any ϵ > 0 we have a set Eϵ with μ(E \ E_{ϵ}) < ϵ and 1/|f| > δ_{ϵ} on E_{ϵ} with δ_{ϵ} = 1/(N + δ).

(2) Let f_{n} : E → R^{∗} be a sequence of measurable functions with f_{n} → f a.e on E. Suppose that |f_{n}| > 0 and |f| > 0 a.e. As above, we may suppose without loss of generality that f_{n}, f are finite a.e. Let ϵ > 0 be arbitrary. Then by (1) above, there exists a set E_{ϵ} ⊂ E and δ_{ϵ} > 0 with μ(E \ E_{ϵ}) < ϵ/3 and |f| > δ_{ϵ} on E_{ϵ}.

Now by the Egorov Theorem, there exists a set E^{(0)} ⊂ E_{ϵ} with μ(E_{ϵ} \ E^{(0)}) < ϵ/3 and f_{n} → f uniformly on E_{ϵ}. Thus, there exists a positive N such that for all n ≥ N we have |f_{n} − f| < δ_{ϵ}/3 on E^{(0)}, so that for all n ≥ N we have |f_{n}| > δ_{ϵ}/3.

Next, we may repeatedly apply problem (1) above to the functions f_{i} for 1 ≤ i < N to obtain sets E^{(i)} ⊂ E^{(i−1)} for i = 1 … N − 1 and positive numbers δ_{i} with μ(E^{(i)} \ E^{(i−1)}) < ϵ/3N and |f_{i}| > δ_{i} on E^{(i)}. Let δ = min{δ_{ϵ}/3, δ_{1}, …, δ_{N}}. Then δ > 0 and by construction we have |f_{i}| > δ on E^{(N)}. Moreover, we have μ(E \ E^{(N)}) ≤ ϵ/3 + ϵ/3 + N ⋯ ϵ/3n = ϵ, as required.

(3) Let f_{n} : E → R^{∗} be a sequence of measurable functions and suppose that f_{n}(x) is finite a.e. on E. Let E_{n,m} = {x : f_{n}(x) ≥ m}. By hypothesis, lim_{m→∞} μ(E_{m,n}) = 0 for all n. Thus, for each n there exists a positive integer m_{n} such that for all m ≥ m_{n} we have μ(E_{m,n}) < 1/2^{n}. Let k_{n} = 2^{n}m_{n}, and let F ⊂ E be the set F = {x : lim_{n→∞} |f_{n}(x)/k_{n}| > 0}. Define F_{n} = {x : |f_{n}(x)/k_{n}| > 2^{−n}}. Observe that x ∈ F implies that x ∈ F_{n} for all sufficiently large n, and hence that x ∈ ⋂_{n=1}^{∞} ⋃_{j=n}^{∞} F_{j}, i.e. F ⊆ lim sup F_{n}. On the other hand F_{n} = {x : |f_{n}(x)/k_{n}| > 2^{−n}} = {x : |f_{n}(x)| > m_{n}} so that μ(F_{n}) < 2^{−n}, and hence ∑_{n} μ(F_{n}) = 1 < ∞. It follows that μ(lim sup F_{n}) = 0 and hence μ(F) = 0 whence f_{n}/k_{n} → 0 a.e.

(4) Recall the definition

∫_{X} f dμ = sup {∫_{X} φ dμ : φ is simple} .

Set

φ_{α}(x) = \begin{cases}α & if f(x) > α \\ 0 & otherwise \end{cases} .

Certainly, φ is simple and we have f ≥ φ on X by construction and since f ≥ 0 on X. It follows that

∫_{X} f dμ ≥ ∫_{X} φ dμ = αμ{x : f(x) > α}.

Chebyschev’s inequality now follows.

(5) Set A_{j} = ⋃_{a=0}^{2^{j}−1}[2a/2^{j+1}, (2a + 1)/2^{j+1}) and B_{j} = ⋃_{a=0}^{2^{j}−1}[(2a + 1)/2^{j}, (2a + 2)/2^{j}). Observe that for each j we have

R_{j}(x) = \begin{cases}−1 & if x ∈ A_{j} \\ 1 & if x ∈ B_{j} \end{cases} .

(a) (i) From the above description, since A_{j} ∪ B_{j} = [0, 1], we see that R_{j}^{1} = 1 on [0, 1] and hence

∫_{0}^{1} R_{j}^{2} dμ = 1.

(ii) Without loss of generality, suppose that j < k. Then for each a, we have

∫_{a/2^{j}}^{(a+1)/2^{j}} R_{j}R_{k} dμ = ∑_{b=2^{k−j}a}^{2^{k−j}(a+1)−1} ∫_{b/2^{k}}^{(b+1)/2^{k}} R_{j}R_{k} dμ = ∑_{b=2^{k−j}a}^{2^{k−j}(a+1)−1}(−1)^{b+a} = 0.

It follows that

∫_{0}^{1} R_{j}R_{k} dμ = 0

for j ≠ k. (iii) We have

∫_{0}^{1}(R_{1} + R_{2} + ⋯ + R_{n})^{2} dμ = ∫_{0}^{1} ∑_{j=1}^{n} R_{j}^{2} + 2 ∑_{1≤j<k≤n} R_{j}R_{k} dμ

= ∑_{j=1}^{n} ∫_{0}^{1} R_{j}^{2} dμ + 2 ∑_{1≤j<k≤n} ∫_{0}^{1} R_{j}R_{k} dμ

= n,

where we have used parts (i) and (ii).

(b) Since (R_{1} + ⋯ + R_{n})^{2} ≥ 0 on [0, 1], we have by problem 4 that

μ(\{x ∈ X : (R_{1} + ⋯ + R_{n})^{2} > n^{2}ϵ^{2}\}) ≤ \frac{1}{n^{2}ϵ^{2}} ∫_{0}^{1}(R_{1} + ⋯ + R_{n})^{2} dμ = \frac{1}{nϵ^{2}}

by part (a), (iii).

(c) It follows from part (b) that \frac{R_{1}+···+R_{n}}{n} tends to 0 in measure as n → ∞; indeed, if it did not,

we would have that

μ({x ∈ X : R_{1} + ⋯ + R_{n} > nϵ})

does not tend to 0, and hence that

μ({x ∈ X : (R_{1} + ⋯ + R_{n})^{2} > (nϵ)^{2}})

does not tend to 0 as n → ∞, and this contradicts part (b).

(d) On the other hand, the set

S = \{x ∈ (0, 1) : \frac{R_{1} + ⋯ + R_{n}}{n} → 0\}

has empty interior. Indeed, let x ∈ S and let ϵ > 0 be arbitrary. Pick m so that 2^{−m} < ϵ and let x = .a_{1}a_{2} … be the binary expansion of x. Now set y = .a_{1}a_{2} … a_{m}10110111011110111110 …, where the pattern of 1’s and 0’s at the end consists of j 1’s followed by a 0 for j = 1, 2, 3, . . .. By construction, y ∈ (x − ϵ, x + ϵ). On the other hand, it is evident by construction that y ∉ S. Since ϵ was arbitrary, it follows that S has empty interior.