Prevent mass from concentrating in tails

Try

A common envelope is convenient but sometimes too restrictive. Instead ask whether any member of a family can hide substantial integral in extremely large values. For f_n=n·1_(0,1/n), the answer is yes: above any fixed threshold, late functions still carry integral one. For g_n=1_(0,1/n), thresholds above one capture nothing. This tail distinction predicts their different L1 behavior.

Predict

Predict whether bounded L1 norms alone prevent concentration. The spike disproves that claim. Now predict whether a uniform bound on a higher moment helps. On the event abs (f)>K, the inequality abs (f)≤abs (f)^p/K^(p−1), for p>1, converts a bounded pth moment into a tail integral tending uniformly to zero.

Experiment

Compare three families on a probability space.

Family	Uniform L1 bound	Large-value tail control
n·1_(0,1/n)	1	Fails
1_(0,1/n)	1	Holds
Any family with integral abs (f)²≤9	At most 3	Tail at most 9/K

Verify the last L1 bound using Cauchy–Schwarz, or its elementary square-integral inequality. Changing a threshold K is not the same as changing the sequence index n: uniformity means one K works for all members.

Understand

On a probability space, a family F of integrable functions is uniformly integrable when sup over f in F of integral abs (f)1_(abs (f)>K) tends to zero as K tends to infinity. This implies bounded L1 norms by splitting at a sufficiently large fixed K. It also implies uniform absolute continuity: the integral over a set A is at most K P(A) plus the uniformly small tail.

Conversely, suppose the L1 norms are bounded by C and integrals are uniformly small on sets of sufficiently small probability. Markov’s inequality gives P(abs (f)>K)≤C/K. For large K, every tail set therefore has sufficiently small probability, proving the tail criterion. This equivalence explains why a condition about large values becomes useful for random exceptional sets.

Vitali’s convergence theorem on this finite measure space states that convergence in probability to a measurable f and uniform integrability imply L1 convergence. First choose a subsequence converging almost surely: pick indices whose probabilities of error greater than 2^-j are below 2^-j, and use the summable union-tail bound. Fatou and the uniform L1 bound show that the limit f is integrable. Its integral is absolutely continuous on small sets by truncation.

For η>0, let A_n be where abs (f_n−f)>η. Its probability tends to zero. The difference integral outside A_n is at most η, while inside it is at most the integrals of abs (f_n) and abs (f). Uniform small-set control makes both arbitrarily small. Then let η decrease to zero. Conversely, L1 convergence gives uniform small-set control by separating finitely many early functions from the close-to-f tail, and bounded norms; the equivalence above proves uniform integrability.

Apply

Apply the theorem to a family with uniformly bounded pth moments, p>1, and convergence in probability. Explain why the moment bound supplies the missing uniform integrability. On an infinite-measure space, the amplitude condition alone is insufficient: (1/n)1_[0, n] converges in measure to zero and has no high-value tails, yet every integral is one. Spatial escape needs additional control.

Continue

Domination is a sufficient route to uniform integrability, not its definition. Keep the finite-measure setting visible when using this version of Vitali. In later probability lessons the same criterion upgrades convergence in probability to convergence of expectations and clarifies when limiting risk calculations are justified.

Continue to Build a product measure from rectangles.