Answer: Lebesgue Differentiation: Failure Everywhere?

Key Idea / Intuition

The Lebesgue Differentiation Theorem is rock-solid: for any $L^1_{\mathrm{loc}}$ function, the averaging limit equals $f(x)$ at almost every point — no measurable function can escape this. The most extreme example of "failure everywhere" is simply impossible. However, the theorem leaves open a measure-zero set of exceptional points, and with a clever construction you can make the limit exist but differ from $f$ on a measure-zero set (e.g., at a jump discontinuity, the limit is the average of left and right limits). The bonus reveals a genuine subtlety: non-symmetric intervals can break the theorem even at Lebesgue points.

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Formal Proof / Solution

Part 1: Cannot Fail Everywhere

Claim: No bounded measurable $f$ can have its Lebesgue averages disagree with $f(x)$ at every $x$.

Proof: By the Lebesgue Differentiation Theorem, for any $f \in L^1_{\mathrm{loc}}(\mathbb{R})$,

$\frac{1}{2r}\int_{x-r}^{x+r} f(t)\,dt \to f(x) \quad \text{for a.e. } x.$

Since "almost every" means the exceptional set has Lebesgue measure zero, and $\mathbb{R}$ has infinite measure, the set where the limit equals $f(x)$ is co-null (complement has measure zero). In particular, it is dense (in fact full measure). So failure at every point is impossible. $\blacksquare$

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Part 2: The Extremal Example — Failure on a Measure-Zero Set

Consider $f = \mathbf{1}_{[0,\infty)}$, the Heaviside step function. Then:

• For $x > 0$: the average over $[x-r, x+r]$ is $1$ for small $r$, and $f(x) = 1$. ✓
• For $x < 0$: the average is $0$ for small $r$, and $f(x) = 0$. ✓
• For $x = 0$:

$\frac{1}{2r}\int_{-r}^{r} \mathbf{1}_{[0,\infty)}(t)\,dt = \frac{1}{2r} \cdot r = \frac{1}{2} \neq f(0) = 1.$

So the limit exists at $x = 0$ but equals $\tfrac{1}{2} \neq f(0)$. This is the classic example: at a jump discontinuity, the symmetric averaging limit gives the midpoint of the jump, not the function value.

The takeaway: the Lebesgue Differentiation Theorem guarantees correctness a.e., but the function's value at a single point (a measure-zero set) is invisible to the integral — you can freely redefine $f$ on a null set without changing any integral. The theorem recovers the equivalence class representative in a canonical sense.

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Bonus: Non-Symmetric Intervals Can Fail at Lebesgue Points

Replace $[x-r, x+r]$ with $[x, x+r]$ (one-sided average). Then the one-sided derivative

$\lim_{r\to 0} \frac{1}{r}\int_x^{x+r} f(t)\,dt$

still equals $f(x)$ at a.e. $x$ (by the same theorem applied to one-sided balls).

But consider a genuinely non-symmetric family of intervals, e.g., $[x - r^2, x + r]$. The ratio of left-to-right length tends to $0$, so effectively this is a one-sided limit. For the Heaviside function at $x = 0$:

$\frac{1}{r^2 + r}\int_{-r^2}^{r} \mathbf{1}_{[0,\infty)}\,dt = \frac{r}{r^2 + r} = \frac{1}{1+r} \to 1 = f(0^+),$

which now recovers the right-hand limit, not the symmetric average $\tfrac{1}{2}$.

More dramatically, using the Busemann–Feller–Morse theorem, there exist differentiation bases (families of sets shrinking to a point) for which the differentiation theorem fails even for continuous functions — the geometry of the sets matters deeply. This is why the theorem is stated for balls (or nicely shaped sets with bounded eccentricity).

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Summary Table

| Scenario | Does limit $= f(x)$ a.e.? | |---|---| | Symmetric intervals $[x-r,x+r]$ | ✓ Yes (LDT) | | One-sided intervals $[x, x+r]$ | ✓ Yes (LDT) | | Badly shaped/non-symmetric basis | ✗ Can fail | | Failure at every point | ✗ Impossible |