Math 597. Homework 7 Bryden Cais

(1) (a) Define E_{j}(η) = {x : |f_{j}| > η} and F_{j}(η) = {x : |f_{j}| ≤ η}. Obviously E_{j}(η) ∪ F_{j}(η) = X for any j. In one direction, suppose that μ(E_{j}(η)) → 0 as j → ∞ for any fixed η > 0. Then we have

∫_{X} \frac{|f_{j}|}{1 + |f_{j}|} dμ ≤ ∫_{E_{j}(η)} dμ + ∫_{F_{j}(η)} |f_{j}| dμ,

since we have the obvious inequality |f_{j}|/(1 + |f_{j}|) ≤ min{1, |f_{j}|}. Since |f_{j}| ≤ η on F_{j}(η), it follows that we have the bound

∫_{X} \frac{|f_{j}|}{1 + |f_{j}|} dμ ≤ μ(E_{j}(η)) + ημ(F_{j}(η))

for each η > 0 and all j. By hypothesis μ(E_{j}(η)) → 0 (equivalently μ(F_{j}(η)) → μ(X) < ∞) so that

lim_{j→∞} ∫_{X} \frac{|f_{j}|}{1 + |f_{j}|} dμ ≤ ημ(X).

Since this holds for any η > 0 and μ(X) is finite, it follows that

∫_{X} \frac{|f_{j}|}{1 + |f_{j}|} dμ → 0.

In the other direction, let 0 < η < 1 be arbitrary. Then it is not difficult to see that

\{x : |f_{j}(x)| > \frac{η}{1 − η} \} = \{x : \frac{|f_{j}(x)|}{1 + |f_{j}(x)|} > η\}.

Moreover, η → 0 if and only if η/(1 − η) → 0. Since we may without loss of generality restrict to only considering η with 0 < η < 1, it follows that |f_{j}| tend to 0 in measure if and only if |f_{j}|/(1 + |f_{j}|) tend to 0 in measure. Now using Chebyschev’s inequality from Homework 6, we have

μ(\{x : |f_{j}|/(1 + |f_{j}|) > η\}) ≤ \frac{1}{η} ∫_{X} \frac{|f_{j}|}{1 + |f_{j}|} dμ.

Since the integral tends to 0 by hypothesis, we see that |f_{j}|/(1 + |f_{j}|) tends to 0 in measure, and hence |f_{j}| tends to 0 in measure. This completes the proof.

(b) Let X = R and f_{j}(x) = \frac{1}{j}. Then we have

μ(\{x : f_{j}(x) > η\}) = \begin{cases}∞ & if η < 1/j \\ 0 & otherwise \end{cases} .

Since for any η > 0 there exists some j > 0 with 1/j < η, it follows that f_{j} tends to 0 in measure. On the other hand,

∫_{R} \frac{|f_{j}|}{1 + |f_{j}|} dμ = ∫_{R} \frac{1}{j + 1} dμ = ∞

for all j. It follows that the integral does not tend to 0 and thus part (a) can fail if μ(X) is not finite.

(2) Recall that the Lebesgue-Stieljes measure ν examined in Problem 3 of Homework 4 satisfies ν(C) = ν([0, 1]) = 1 and ν([0, 1] \ C) = 0, where C is the usual cantor set (of measure 0) on the interval [0, 1]. If ν were induced as

ν(E) = ∫_{E} g dμ

for some nonnegative measurable function g with μ being Lebesgue measure, then we would necessarily have

1 = ν([0, 1]) = ∫_{[0,1]} g dμ = ∫_{[0,1]\C} g dμ = ν([0, 1] \ C) = 0,

since we can “throw away” any set of measure 0 from the integral. This is a contradiction so that there is no such g inducing ν as above.

(3) Since the a_{j} are positive, the Monotone Convergence Theorem tells us that

∫_{0}^{1} ∑_{j=1}^{∞} \frac{a_{j}}{\sqrt{|x − r_{j}|}} dμ = ∑_{j=1}^{∞} ∫_{0}^{1} \frac{a_{j}}{\sqrt{|x − r_{j}|}} dμ.

Now observe that

∫_{0}^{1} \frac{a_{j}}{\sqrt{|x − r_{j}|}} dμ = ∫_{0}^{r_{j}} \frac{a_{j}}{\sqrt{r_{j} − x|}} dμ + ∫_{r_{j}}^{1} \frac{a_{j}}{\sqrt{x − r_{j}}} dμ

= −2a_{j}\sqrt{r_{j} − x} ]_{0}^{r_{j}} + 2a_{j}\sqrt{x − r_{j}} ]_{r_{j}}^{1}
= 2a_{j}\sqrt{r_{j}} + 2a_{j}\sqrt{1 − r_{j}}
≤ 4a_{j},

so that

∑_{j=1}^{∞} ∫_{0}^{1} \frac{a_{j}}{\sqrt{|x − r_{j}|}} dμ

is finite. If ∑_{j=1}^{∞} \frac{a_{j}}{\sqrt{|x−r_{j}|}} were not finite a.e. then the integral

∫_{0}^{1} ∑_{j=1}^{∞} \frac{a_{j}}{\sqrt{|x − r_{j}|}} dμ

would not be finite; by our considerations (with MCT) above, this is a contradiction.

(4) Suppose that f(x) and |x|f(x) are integrable on R.

(a) Observe that for fixed y we have

f(x)|y − x| = \begin{cases}f(x)y − xf(x) & if y − x ≥ 0 \\ xf(x) − f(x)y & otherwise \end{cases} .

Since f(x), |x|f(x) are integrable, so are ∓yf(x) ± xf(x) since y is constant. It follows that f(x)|y − x| is integrable.

(b) Set

ϕ_{h}(x) = f(x)\frac{|y + h − x| − |y − x|}{h}

and observe that

|ϕ_{h}(x)| ≤ |f(x)|\frac{|h| + |y − x| − |y − x|}{|h|} = |f(x)|,

so that for all h, we have a dominating function |f(x)| with |ϕ_{h}| ≤ |f|. It follows from the Dominated Convergence Theorem that

h′(y) := lim_{h→0} ∫_{R} ϕ_{h} dμ = ∫_{R} lim_{h→0} ϕ_{h} dμ.

On the other hand, we have

lim_{h→0} ϕ_{h} = f(x) lim_{h→0} \frac{|y − x + h| − |y − x|}{h} = f(x)sgn(y − x).

The proposed equality follows.