Integration Theory / Mathematical Analysis (5B1452/MA429)

Solutions to homework assignment # 2

1. a. Not integrable. In fact, for any n one has ∫_{2πn}^{2πn+π} f(x)dx = 2, so

∫_{R} |f(x)|dx ≥ ∑_{n=1}^{∞} ∫_{2πn}^{2πn+π} f(x)dx = ∑_{n=1}^{∞} 2 = +∞ ,

and f cannot be integrable. b. Not integrable. By the Lebesgue monotone convergence Theorem,

∫_{[0,1]} f(x)dx = lim_{n→∞} ∫_{1/n}^{1} f(x)dx = lim_{n→∞}(n − 1) = ∞ ,

so the integral exists and is equal to +∞, but f is not integrable. c. Integrable. Same reasoning shows, that

∫_{[1,∞]} f(x)dx = lim_{n→∞} ∫_{1}^{∞} f(x)dx = lim_{n→∞}(1 − \frac{1}{n}) = 1 < +∞ ,

so f(x) is integrable since it is measurable and non-negative.

2. a. By a classical inequality, 1 + nx ≤ (1 + x)^{n}, so we deduce that sunctions f_{n}(x) := \frac{1+nx}{(1+x)^{n}} have an integrable majorant on [0, 1], namely 0 ≤ f_{n}(x) ≤ 1. Also f_{n} converges pointwise to f(x), which is 0 for non-zero x and 1 otherwise: f(0) = 1. Therefore by the Lebesgue bounded convergence Theorem,

lim_{n→∞} ∫_{0}^{1} \frac{1 + nx}{(1 + x)^{n}} dx = ∫_{[0,1]} f(x)dx = 0 .

b. Define functions f_{n}(x) on [0, ∞) by f_{n}(x) := (1 + x/n)^{n}e^{−2x}, if x ∈ [0, n], and f_{n}(x) := 0 otherwise. By a classical ineqaulity (1 + \frac{1}{y})^{y} < e. Plugging in y = n/x, we obtain 0 ≤ f_{n}(x) = (1 + x/n)^{n}e^{−2x} = (1 + x/n)^{(n/x)x}e^{−2x} ≤ e^{x}e^{−2x} = e^{−x}, the latter clearly integrable. Moreover, by an equally classical limit, lim_{n→∞}(1 + x/n)^{n} = e^{x}, and hence f_{n}(x) conveges pointwise to e^{−x}.

Thus we can apply the Lebesgue bounded convergence Theorem, and write

lim_{n→∞} ∫_{0}^{n}(1 + x/n)^{n}e^{−2x}dx = lim_{n→∞} ∫_{[0,∞]} f_{n}(x)dx = ∫_{[0,∞]} e^{−x}dx = −e^{−x}|_{x=0}^{∞} = 1 .

3. a. If g_{1} is integrable, then we can apply the Lebsgue bounded convergence Theorem with majorant g_{1} (since 0 ≤ g_{n} ≤ g_{1}), and obtain integrability of g_{n}, g, and desired convergence of integrals. b. On the real line with Lebesgue measure, take g_{n} := χ_{[n,∞)}, they clearly converge to g(x) = 0, without convergence of integrals.

4. If we prove this statements for (non-negative) functions f^{+} and f^{−}, statements for f clearly follow, so we can assume from the beginning that we deal with a non-negative function f.

Consider (biinfinite) sequence of functions {g_{n}}_{n=−∞}^{+∞}, defined by g_{n}(x) := f(x + n). They are clearly measurable, and denoting g := ∑_{n=−∞}^{+∞} g_{n} we can apply the Theorem about integrals of series (for the interval [0, 1]):

∫_{0}^{1} g(x)dx = ∑_{n=−∞}^{+∞} ∫_{0}^{1} g_{n}(x)dx = ∑_{n=−∞}^{+∞} ∫_{0}^{1} f(x + n)dx

= ∑_{n=−∞}^{+∞} ∫_{n}^{n+1} f(x)dx = ∫_{−∞}^{+∞} f(x)dx < +∞ .

Thus, function g is integrable on [0, 1], and part b. follows. Also it implies, that g is finite a.e.. Therefore the series in question converges for a.e. x in [0, 1] and hence for a.e. x on the real line (the sum of this series is periodic, g(x) = g(x + 1)).

5. a. Function f is equal to the function

g(x) := ∑_{n=1}^{∞} nχ_{(1/(n+1),1/n]}(x) ,

changed on the set of rational numbers to be equal to zero there. Since g is a some of measurable functions (in fact, chracteristic functions of measurable sets), it is measurable. But f = g almost everywhere, thus (Lebesgue measure is complete, so all nullsets are measurable) f is also measurable. b. Functions f and g have the same integrals (since f = g a.e.), so we can write

∫_{0}^{1} f(x)dx = ∑_{n=1}^{∞} ∫_{(1/(n+1),1/n]} f(x)dx = ∑_{n=1}^{∞} ∫_{(1/(n+1),1/n]} ndx

= ∑_{n=1}^{∞}(1/n − 1/(n + 1))n = ∑_{n=1}^{∞} 1/(n + 1) = +∞ .