Lebesgue Measure and Integration

With AC, |S × S| = |S| for every infinite S. (1), (2) fix a specific cardinality; (4) contradicts Cantor's theorem.

Every subset of X is the zero set of its indicator-complement (f = 1 off the subset, 0 on it), so Y is the power set of X, with $2^{|X|}$ > |X| elements.

The removed intervals have total length $\sum 2^n$⁻${}^1/3^n=1$, so C has Lebesgue measure 0; yet the ternary-expansion coding (digits 0 and 2) puts C in bijection with {0, 1}${}^{\mathbb{N}}$, which is uncountable. Small in measure, large in cardinality.

$\int f_n=1$ for every n, so Fatou is strict and DCT cannot apply — no integrable g dominates all $f_n$.

For x > 0 the terms vanish once n > 1/x, and $f_n(0)=0$ always, giving (1). But $\int f_n=n\cdot (1/n)=1$ for every n, so (2) fails. (3) fails: a dominating function would need $g(x)\ge su{p}_n{f}_n(x)\approx 1/x$ near 0, which is not integrable — exactly why the limit of the integrals (1) differs from the integral of the limit (0). (4) Fatou reads 0 ≤ 1, a strict inequality, which the lemma permits.

Finite measure reverses the inclusion (Hölder); for sequences the inclusion goes up. On $\mathbb{R}$ neither holds: $1/\sqrt{x}\cdot 1_{(0,1)}$ is in $L^1$ not $L^2$, and 1/x·$1_{(1,\infty )}$ is in $L^2$ not $L^1$.

On a finite measure space Cauchy–Schwarz gives $\int$|f| $=\int$|f|·1 ≤ ‖f‖${}_2$‖1‖${}_2=$ ‖f‖${}_2$, proving (1) and (3) at once. (2) fails: f(x) = $x^{-1/2}$ on (0,1) is integrable but its square is not. (4) fails because the measure is infinite: f(x) = 1/x on $(1,\infty )$ is in $L^2$ but not in $L^1$ — the inclusion direction depends entirely on finiteness of the measure.