Compare almost uniform and nearly continuous behavior

Try

On [0,1], the functions 1_(0,1/n) converge pointwise to zero but never uniformly on the whole interval. Remove a short interval (0,δ), however, and every sufficiently late function is identically zero on the remainder. This suggests that pointwise convergence on a finite-measure space can be made uniform after sacrificing a small set. The small set need not be null.

Predict

Predict whether one exceptional set works for every requested error budget. Egoroff’s theorem lets the set depend on that budget. Predict also whether finite total measure matters. Translating unit intervals on the whole real line provide a counterexample: deleting less than one unit of measure cannot eliminate all the late peaks. Distinguish a small measure of bad points from a small maximum error.

Experiment

For a tolerance sequence 1/k, build a separate bad tail at each level.

Tolerance	Bad tail	Allowed measure
1	E_(N₁,1)	ε/2
1/2	E_(N₂,2)	ε/4
1/3	E_(N₃,3)	ε/8

The indices need not be equal or even chosen minimally. Their union costs at most ε. On its complement, choose the row whose tolerance is below your requested error, then use that row’s tail index. This order of choices proves uniform convergence.

Understand

Suppose μ(X)< infinity and real measurable f_n→f almost everywhere. Let E_(N, k) be the union over n≥N of the sets where abs (f_n−f)>1/k. As N increases, these sets decrease to a null set because convergence eventually meets each fixed tolerance. Continuity above applies since the ambient measure is finite. Choose N_k so that μ(E_(N_k, k))<ε2^-k. Outside the union E of these selected bad tails, all n≥N_k satisfy abs (f_n−f)≤1/k. Since k can be chosen arbitrarily large, convergence is uniform there. This proves Egoroff’s theorem.

Lusin’s theorem concerns measurable functions rather than sequences: a finite-valued Lebesgue measurable function on a bounded closed interval is continuous relative to a compact subset whose omitted measure is arbitrarily small. Here is a construction. Approximate the function pointwise by finite-valued simple functions, using truncation and range subdivision for signed values. For each simple function, choose compact subsets inside its finitely many level sets by inner regularity, spending an assigned small error budget. On their finite union, the simple function is relatively continuous because distinct compact pieces are separated.

Intersect these compact good sets over the sequence; the total omitted measure is bounded by the sum of budgets. Egoroff supplies a further measurable good set of large measure where the simple approximations converge uniformly. Choose a compact subset inside that set, again by regularity, and intersect it with the previous compact intersection. On the resulting compact set all approximants are continuous and converge uniformly, so their limit is continuous. Choose the budgets to total less than ε. Null exceptional points where the original function is not finite can also be discarded.

Apply

Neither theorem changes the underlying function values on the retained set. Egoroff improves convergence; Lusin improves relative continuity. For a step function, remove a tiny neighborhood of its jump and verify the compact-set conclusion directly. Explain why this does not prove continuity at the deleted jump, nor supply one universal compact set for every error budget.

Continue

These results rely on different ingredients: finite measure for Egoroff and Lebesgue regularity for Lusin’s compact approximation. Keep them separate in your proof notes. Next, control integrals uniformly over small sets rather than insisting on pointwise domination; that will lead to a more general convergence criterion.

Continue to Prevent mass from concentrating in tails.