Recover an integrable function from local averages

Try

Average a locally integrable signal over a short interval around a point. For continuous signals the averages approach the point value, but measurable signals can have dense discontinuities. The differentiation theorem says that local averages still recover the value almost everywhere. Its proof controls where averaging can amplify an approximation error.

Predict

Predict whether the statement holds at every point. A jump gives an immediate warning: symmetric averages may approach the midpoint of the two sides rather than the assigned point value. Predict whether a small global L1 error guarantees a small local average everywhere. A concentrated spike disproves that; instead we bound the measure of locations where amplification is large.

Experiment

For f(x)=x² at x=0, the centered average over [−r, r] is r²/3.

Radius r	Average	Point value
1	1/3	0
1/2	1/12	0
1/4	1/48	0

Now draw a narrow high spike with the same total integral as a low wide bump. Identify the centers whose averaging intervals can capture substantial mass. The proof bounds their total length, not each local average individually.

Understand

For f in L1 of the real line, define the centered maximal function Mf (x) as the supremum over r>0 of the average of abs (f) on (x−r, x+r). Each fixed-radius average is continuous in x: moving the interval changes it only on a set of small measure, and the integral of an L1 function is absolutely continuous over such sets. Therefore the strict level set where Mf>α is open.

We prove m(Mf>α)≤3 norm (f)_1/α. Take a compact subset K of this level set. For each point choose a centered witness interval whose average exceeds α, and use compactness to select a finite cover of K. Select a longest remaining interval and discard all intervals meeting it, repeating until none remain. The selected intervals are disjoint. Any discarded interval has no greater length and meets its selected partner, so it lies in that partner’s concentric triple. Thus m(K) is at most three times the sum of selected lengths, bounded by 3 norm (f)_1/α because their integrals are disjoint. Inner regularity of the open level set proves the estimate.

Now approximate f in L1 by a continuous compactly supported g and put h=f−g. At each point where f is finite, the limsup of the local average of abs (f(y)−f(x)) is at most Mh (x)+abs (h(x)); the continuous g contribution tends to zero. The set where this limsup exceeds 2α is contained in the union where Mh>α or abs (h)>α. The maximal estimate and Markov bound its measure by 4 norm (h)_1/α. Density lets this norm tend to zero while the bad set for f stays fixed. Taking countably many positive α tending to zero proves that the local mean oscillation vanishes almost everywhere. This is the Lebesgue differentiation theorem.

Apply

For locally L1 functions, truncate to successively larger bounded intervals and apply the theorem away from their boundaries. A countable union of exceptional null sets remains null, proving the local version. For an indicator, the theorem says almost every point of a measurable set is a density-one point, even when the set has no interior. Explain why density one is weaker than containing a neighborhood.

Continue

The maximal estimate turns norm approximation into an almost-everywhere local statement. The constant three is sufficient here, not asserted optimal. Next, connect differentiation of integrals to absolute continuity and identify precisely why continuity alone does not recover a function from its derivative.

Continue to Identify exactly which functions are integrals of derivatives.