Construct limits of Cauchy sequences

Try

Suppose successive approximations differ in Lp norm by at most 1/2, 1/4, 1/8 and so on. The total possible movement is bounded. To prove an actual limit exists, however, bounded movement is not enough by itself: construct a function and show the approximations converge to it in the claimed norm. This construction is the substance of completeness.

Predict

Predict whether a norm-Cauchy sequence must converge at every point. Representatives may be changed arbitrarily on null sets, so an every-point assertion cannot be intrinsic to Lp. The correct proof produces convergence almost everywhere for a selected subsequence, then norm convergence for the original sequence. Do not confuse those two conclusions.

Experiment

Record the remaining movement after successive stages.

Starting difference index	Sum of remaining bounds
1	1
2	1/2
3	1/4
5	1/16

These bounds come from a geometric series, not observed pointwise errors. Use them to specify a subsequence index that guarantees a desired norm tolerance. Then explain how the original Cauchy condition controls indices between the selected terms.

Understand

Let f_n be Cauchy in Lp, 1≤p< infinity. Choose a subsequence h_k with norm (h_(k+1)−h_k)_p≤2^-k. Write d_k=h_(k+1)−h_k. Minkowski bounds the Lp norm of each partial sum of abs (d_k) by one. These sums increase pointwise. Monotone convergence applied to their pth powers shows that g=Σabs (d_k) belongs to Lp and is finite almost everywhere.

Outside a measurable null exceptional set, the series h_1+Σd_k converges absolutely to a finite function h. Define h to be zero on the exceptional set; the resulting representative is measurable. Its absolute value is bounded by abs (h_1)+g, so it belongs to Lp. Fatou applied to finite tails gives norm (h−h_m)_p≤Σ_(k≥m)2^-k, tending to zero. Finally, for large n choose a late subsequence member h_m close both to h and, by the Cauchy condition, to f_n. Minkowski proves convergence of the whole sequence. Thus Lp is complete, a Banach space.

For L infinity use the same subsequence but remove all exceptional null sets where the essential difference bounds fail, also removing a set where the first representative exceeds a finite essential bound. On the complement the series converges uniformly to a bounded measurable limit. The uniform tail estimate proves essential-sup convergence.

For finite p, simple functions of finite-measure support are dense. Truncate f to the set where 1/n≤abs (f)≤n. This support has finite measure because μ(abs (f)≥1/n)≤n^p integral abs (f)^p. Dominated convergence controls the omitted pth-power integral. On the retained support a finite value grid gives an arbitrarily small Lp error. For Lebesgue measure, regularity further approximates indicator sets by continuous compactly supported functions: choose compact K inside open bounded U and a continuous function equal to one on K, zero outside U, between zero and one. The error is supported in U outside K, whose measure can be made small.

Apply

One explicit continuous separator is dist (x, U complement) divided by dist (x, U complement)+dist (x, K), when K is nonempty and compactly contained in U. Explain why its denominator never vanishes and why its support can be kept bounded. Combine this construction with simple-function approximation to prove continuous compactly supported functions are dense for finite p.

Continue

Completeness supplies limits of iterative approximations; density supplies manageable test functions. Neither says continuous functions approximate every L infinity function in essential supremum. The jump-indicator counterexample in practice marks that boundary and prevents an incorrect transfer of a finite-p theorem.

Continue to Separate positive and negative mass.