Real Analysis, Math 821.

Instructor: Dmitry Ryabogin

Assignment VII.

1. Problem 1.

a) Compute the Lebesgue integral (with respect to usual Lebesgue measure) over (0, +∞) from the following functions ℵ) f(x) = e^{−[x]}, ℶ) f(x) = 1/([x + 1][x + 2]), ℷ) 1/[x]!. Here [x] stands for the integer part of x.

b) Find the set of values of α, β, such that f(x) := x^{α} sin x^{β}, x ∈ (0, 1], ℵ) is Lebesgue integrable,

ℶ) is “improper” Riemann integrable, i.e. there exists a limit lim_{ϵ→0} ∫_{ϵ}^{1} f(x)dx.

c) Compute the Lebesgue integral over [0, π/2] from the following functions ℵ) f(x) = sin x, ℶ) f(x) = sin x for x ∈ Q, and f(x) = cos x for x ∈ R \ Q, ℷ) f(x) = sin x for cos x ∈ Q, and f(x) = sin^{2} x for cos x ∈ R \ Q.

2. Problem 2. a) Let φ be an increasing smooth function on [a, b], and let ψ be its inverse on [φ(a), φ(b)]. Prove that

∫_{a}^{b} φ(x)dx = ∫_{φ(a)}^{φ(b)}yψ′(y)dy,

where the integral is understood as a limit of Lebesgue integral sums.

b) Let f(x) ≥ 0 on [a, b], and let m be Lebesgue measure on R^{2}. Prove that

∫_{a}^{b} f(x)dx = m({(x, y) : a ≤ x ≤ b, 0 ≤ y ≤ f(x)}).

c) Let X : μ(X) < ∞, and let f ≥ 0 on X. Prove that f is Lebesgue integrable on X if and only if

∑_{n=0}^{∞} 2^{n}μ({x ∈ X : f(x) ≥ 2^{n}}) < ∞.

3. Problem 3. Let x ∈ [0, 1] be written as ternary fraction x = ∑_{i=1}^{∞} α_{i}/3^{i}, where α_{i} =

0, 1, 2. Assume also that k(x) is the smallest index in this representation for which α_{k} = 1, (if α_{i} ≠ 1 then k(x) = ∞). Define

c(x) := ∑_{i=1}^{k−1} \frac{α_{i}}{2^{i+1}} + \frac{1}{2^{k}} .

Observe that c(x) does not depend on the choice of the ternary representation of x. The function c(x) is called the Cantor ladder, and it is continuous and monotonic (why?). (Observe that c(x) is constant on every interval complement to the Cantor set, and is taking values 1/2^{k}, 3/2^{k}, 5/2^{k},...,(2^{k} − 1)/2^{k} on complemental intervals of the k-th rank).

Compute the following integrals with respect to the measure c (for simplicity I denote it by the same letter), defined on a subring of sets [a, b) ⊂ [0, 1] as c([a, b)) = c(b)−c(a). (It does not really matter here whether we take open, closed, half-open, or half-closed intervals because the function c is continuous).

ℵ) ∫_{0}^{1} x^{k}dc(x),

ℶ) ∫_{0}^{1} e^{x}dc(x),

ℷ) ∫_{0}^{1} sin(πx)dc(x).

Hint. c(x/3) = 1/2c(x), c(2/3 + x/3) = 1/2 + 1/2c(x). Thus,

a_{k} := ∫_{0}^{1} x^{k}dc(x) = ∫_{0}^{1/3}x^{k}dc(x)+∫_{2/3}^{1} x^{k}dc(x) = \frac{1}{2 · 3^{k}} ( ∫_{0}^{1} y^{k}dc(y)+∫_{0}^{1}(2+y)^{k}dc(y)) = ...,

Ψ(a) := ∫_{0}^{1} e^{ax}dc(x) = ∫_{0}^{1/3}e^{ax}dc(x) + ∫_{2/3}^{1} e^{ax}dc(x) = ...

4. Problem 4. Prove the Lebesgue criterion of the Riemann integrability: a bounded f is Riemann integrable on [0, 1] iff it is continuous on [0, 1] almost everywhere.

Hint. Define m(x) = lim_{δ→+0} m_{δ}(x), M(x) = lim_{δ→+0} M_{δ}(x), where

m_{δ}(x) = inf_{x−δ<y<x+δ} f(y), M_{δ}(x) = sup_{x−δ<y<x+δ} f(y).

a) Take a sequence of partions of [0, 1]

0 = x_{0}^{1} < x_{1}^{1} < ... < x_{n_{1}}^{1} = 1,  ...,  0 = x_{0}^{i} < x_{1}^{i} < ... < x_{n_{i}}^{i} = 1,  ...,

such that λ_{i} := max(x_{k+1}^{i} − x_{k}^{i}) → 0, as i → ∞, and define m_{k}^{i} := inf_{[x_{k+1}^{i},x_{k}^{i}]} f(x). Prove that lim_{i→∞} φ_{i}(x) = m(x) almost everywhere, where

φ_{i}(x) = m_{k}^{i}  for x ∈ (x_{k}^{i}, x_{k+1}^{i}),  φ_{i}(x) = 0 for x = x_{0}^{i}, x_{1}^{i}, ..., x_{n_{i}}^{i}.

b) Prove that

lim_{i→∞} ∫_{0}^{1} φ_{i}(x)dx = ∫_{0}^{1} m(x)dx,

and that the corresponding result is true for M.

c) Let s_{i}, S_{i} denote the upper and the lower Darbour sums for the integral of Riemann. Observe that (use the Lebesgue theorem on dominated convergence)

∫_{0}^{1} φ_{i}(x)dx = ∑_{k=0}^{n_{i}−1} ∫_{x_{k}^{i}}^{x_{k+1}^{i}}φ_{i}(x)dx = ∑_{k=0}^{n_{i}−1} m_{k}^{i}(x_{k+1}^{i} − x_{k}^{i}) = s_{i},

∫_{0}^{1} φ̂_{i}(x)dx = ∑_{k=0}^{n_{i}−1} ∫_{x_{k}^{i}}^{x_{k+1}^{i}}φ̂_{i}(x)dx = ∑_{k=0}^{n_{i}−1} M_{k}^{i}(x_{k+1}^{i} − x_{k}^{i}) = S_{i},

and conclude that

lim_{i→∞}(S_{i} − s_{i}) = ∫_{0}^{1}(M(x) − m(x))dx.

d) Observe that

∫_{0}^{1}(M(x) − m(x))dx = 0

iff f is Riemann integrable.