Control limits with an integrable envelope

Try

A pointwise approximation can miss narrow spikes. To recover reliable totals, require one integrable envelope that bounds every approximation. On [0,1], the functions x/(1+x/n) converge to x and are bounded by x itself. Their limiting integral is therefore 1/2. Before invoking a theorem, identify the envelope, its integral and the exceptional set where convergence might fail.

Predict

Compare a shrinking indicator, a tall shrinking spike and a translating indicator. All converge pointwise to zero, but their integrals behave differently. Predict which sequence can fit underneath one integrable envelope. A finite displayed list of functions always admits a finite maximum; the theorem requires one bound for the entire infinite sequence.

Experiment

Live experiment: a pointwise limit can lose integral mass

Predict the area under each graph as n increases. The index selector samples an infinite mathematical sequence; five displayed terms cannot prove its limit. Integrals below are analytic, not estimated from pixels.

Sequence Concentrating spike: n on (0,1/n)Shrinking indicator: 1 on (0,1/n)Escaping interval: 1 on [n,n+1]Monotone ramp: min(nx,1)Index n 124816Probe x 00.06250.1250.512816Reset experiment

The horizontal axis is fixed at [0,1] within this example; the vertical axis stays [0,16] for every example. Jump segments show one-sided heights, not extra function values. Outside the stated support the function is zero. At the selected x=0.125, fₙ(x)=4 and the pointwise limit is 0.

Integral of this term: 1. Integral of the pointwise limit: 0. The integral gap remains one for every index.

For each fixed x>0, eventually 1/n≤x, so fₙ(x)=0; at x=0 it is always zero. Yet height n times width 1/n gives integral one. There is no single integrable dominating function on [0,1]. Indeed any such majorant must dominate the dyadic spikes: on (2⁻⁽ᵏ⁺¹⁾,2⁻ᵏ), its value is at least 2ᵏ, contributing at least 1/2 on each of infinitely many disjoint intervals. The shrinking supports also show failure of uniform integrability.

Compare the two shrinking supports before switching to escaping mass. State the exact missing hypothesis when interchanging limit and integral fails. Settings are temporary; no animation or automatic mastery award is used.

Use the live explorer above to compare exact areas with pointwise readings. The controls show finitely many members and do not prove an infinite-sequence claim. In particular, inspect a fixed x while increasing n, then separately inspect the changing support. These are different observations.

The separate worked example is f_n(x)=x/(1+x/n) on [0,1]. Verify 0≤f_n≤x and pointwise convergence to x. No estimate of the exact intermediate integral is needed to justify the limiting total.

Understand

For a real measurable function f, write f=f⁺−f⁻, where the positive and negative parts are nonnegative and have disjoint supports. Call f integrable when integral abs (f) is finite; then both parts have finite integrals and their difference is unambiguous. A function whose two parts both have infinite integrals has no signed Lebesgue integral under this definition. Cancellation of two infinite totals is not an allowed shortcut.

The dominated convergence theorem assumes measurable f_n and a measurable limit f with f_n→f almost everywhere and abs (f_n)≤g almost everywhere for a single integrable g. The countable union of exceptional null sets is null, and outside it abs (f)≤g as well. All functions involved are therefore integrable.

Apply Fatou to the nonnegative functions g+f_n. Subtracting the finite integral of g gives integral f≤liminf integral f_n. Applying Fatou to g−f_n gives limsup integral f_n≤integral f. The upper and lower limits agree, so the integrals converge. Applying the same conclusion to abs (f_n−f), which tends to zero and is bounded by 2g, proves the stronger statement integral abs (f_n−f)→0. This is convergence in L1.

An envelope controls mass, not merely heights at individual points. The constant one works on the unit interval but fails to be integrable on the whole real line. The concentrating spike n·1_(0,1/n) has bounded integrals yet no integrable common envelope. Thus a uniform bound on integrals is not a substitute for pointwise domination. A later lesson replaces domination by uniform integrability, a more flexible tail condition.

Apply

To differentiate an integral depending on a parameter, apply this theorem to difference quotients, not merely to the original functions. For exp (−tx) on [0,1], fix t and use the one-variable mean value theorem to bound nearby difference quotients by x exp ((abs (t)+1)x). This bound is independent of the small increment and integrable. Explain why a bound depending uncontrollably on the increment would not complete the argument.

Continue

Write a four-line convergence certificate: measure space; pointwise or almost-everywhere limit; common integrable bound; resulting integral or L1 conclusion. If one line fails, inspect concentration, escape or cancellation before trying another theorem. The finite explorer supplies counterexample intuition; the Fatou argument supplies the infinite-sequence proof.

Continue to Compare almost uniform and nearly continuous behavior.