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Spring 2013 - Problem 1

Lebesgue differentiation theorem

Suppose f ⁣:R→R\func{f}{\R}{\R} is bounded, Lebesgue measurable, and

lim⁡h→0∫01∣f(x+h)−f(x)∣h dx=0.\lim_{h\to0} \int_0^1 \frac{\abs{f\p{x + h} - f\p{x}}}{h} \,\diff{x} = 0.

Show that ff is a.e. constant on the interval [0,1]\br{0, 1}.
Solution.

Since ff is bounded, f∈Lloc1(R)f \in L^1_{\mathrm{loc}}\p{\R} so almost every x∈[0,1]x \in \br{0, 1} satisfies

lim⁡h→01h∫xx+hf(t) dt=f(x),\lim_{h\to0} \frac{1}{h} \int_x^{x+h} f\p{t} \,\diff{t} = f\p{x},

by the Lebesgue differentiation theorem, since [x,x+h)\pco{x, x + h} shrinks nicely to xx. Let a<ba < b be two such Lebesgue points and so

∣f(a)−f(b)∣=lim⁡h→0 1h∣∫aa+hf(x) dx−∫bb+hf(x) dx∣=lim⁡h→0 1h∣∫abf(x) dx−∫a+hb+hf(x) dx∣=lim⁡h→0 1h∣∫abf(x) dx−∫abf(x+h) dx∣≤lim⁡h→0 1h∫ab∣f(x+h)−f(x)∣ dx≤lim⁡h→0 1h∫01∣f(x+h)−f(x)∣ dx=0.\begin{aligned} \abs{f\p{a} - f\p{b}} &= \lim_{h\to0}\,\frac{1}{h} \abs{\int_a^{a+h} f\p{x} \,\diff{x} - \int_b^{b+h} f\p{x} \,\diff{x}} \\ &= \lim_{h\to0}\,\frac{1}{h} \abs{\int_a^b f\p{x} \,\diff{x} - \int_{a+h}^{b+h} f\p{x} \,\diff{x}} \\ &= \lim_{h\to0}\,\frac{1}{h} \abs{\int_a^b f\p{x} \,\diff{x} - \int_a^b f\p{x + h} \,\diff{x}} \\ &\leq \lim_{h\to0}\,\frac{1}{h} \int_a^b \abs{f\p{x + h} - f\p{x}} \,\diff{x} \\ &\leq \lim_{h\to0}\,\frac{1}{h} \int_0^1 \abs{f\p{x + h} - f\p{x}} \,\diff{x} \\ &= 0. \end{aligned}

Thus, f(a)=f(b)f\p{a} = f\p{b} for almost any two a<ba < b, which completes the proof.