$f$ Lebesgue integrable, showing $f$ is constant almost everywhere

Question

$f\in L^1([a,b])$ satisfies $\lim_{h\rightarrow 0}\frac{1}{h}\int_{a}^{h}|f(t+h)-f(t)|dt = 0$ then $f$ is constant almost everywhere.

I tried using the mean value theorem, to somehow get that $f'=0$ almost everywhere, but can't seem to prove it is.

Answer 1

I'll assume the hypothesis is

$$\lim_{h\to 0}\frac{1}{h}\int_c^{d} |f(t+h)-f(t)|\, dt = 0$$

whenever $a <c < d<b.$ Define $F(x) =\int_a^x f.$ Then $F'(x) =f(x)$ for a.e. $x\in[a,b].$ Fix an $x_0\in (a,b)$ where $F'(x_0) =f(x_0).$ Let $x\in (a,b).$ Then

$$F(x+h) - F(x) = \int_{x_0}^{x+h} f - \int_{x_0}^{x} f= \int_{x_0}^{x_0+h}f + \int_{x_0+h}^{x+h}f - \int_{x_0}^{x} f$$ $$= \int_{x_0}^{x_0+h}f +\int_{x_0}^{x} (f(t+h) - f(t))\, dt.$$

Now divide by $h.$ We then have

$$\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_{x_0}^{x_0+h}f +\frac{1}{h}\int_{x_0}^{x} (f(t+h) - f(t))\, dt.$$

As $h\to 0,$ the first term on the right $\to f(x_0)$ because $F'(x_0) = f(x_0).$ The second term $\to 0$ by the main hypothesis in the problem. Thus $F'(x) = f(x_0).$

Now note what we have shown: For any $x_0 \in (a,b)$ where $F'(x_0)$ exists, $F'(x) = f(x_0)$ for all $x=(a,b).$ Thus $f$ must be constant on the set of such $x_0.$ Because $F'$ exists a.e., we have shown there is a set of measure $0$ off of which $f$ is constant. That is the desired result.