Proving a measurable function is constant almost everywhere

If $f:\mathbb R\to\mathbb R$ is measurable and bounded and $$\lim_{r\to0}\int_0^1\frac{|f(x+r)-f(x)|}rdx=0$$ then $f$ is constant for a.e. $x\in[0,1]$.

I'm not sure how to go about proving that something's constant almost everywhere. One thing I tried was assuming $f(x)=\lim\frac1{|B(x,r)|}\int_{B(x,r)}f(y)dy$ (which is true almost everywhere by the Lebesgue differentiation theorem) and then proving that $f$ is a constant everywhere but this didn't work out.

• Should $\frac{1}{B(x,r)}$ be $\frac{1}{\mu(B(x,r))} = \frac{1}{2r}$? – Omnomnomnom Jun 24 '17 at 23:20
• @Omnomnomnom thanks – alex-tang Jun 24 '17 at 23:21
• Perhaps it is useful to write $$f(x_2) - f(x_1) = \lim_{r \to 0} \frac{1}{|B(x_1,r)|} \int_{B(x_1,r)} (f(y + (x_1 - x_2)) - f(y))\,dy$$ and WLOG take $x_1 = 0$. – Omnomnomnom Jun 24 '17 at 23:29

Since $f$ is bounded and measurable,

$$F(x) = \int_0^x f(t) \,dt$$

is differentiable and $F'(x) = f(x)$ almost everywhere.

We have

\begin{align}F(x+r) &= \int_0^{x+r}f(t) \, dt \\ &= \int_{0}^{r}f(t) \, dt +\int_r^{x+r} f(t) \, dt \\ &= \int_{0}^{r}f(t) \, dt + \int_{0}^{x}f(t+r) \, dt \end{align}

and

$$\tag{*}\frac{F(x+r) - F(x)}{r} = \frac{1}{r} \int_0^r f(t) \, dt + \frac{1}{r} \int_0^x[f(t+r) - f(t)] \, dt.$$

Note that

$$\left| \frac{1}{r} \int_0^x[f(t+r) - f(t)] \, dt\right| \leqslant \int_0^x\frac{|f(t+r) - f(t)|}{r} \, dt \leqslant \int_0^1\frac{|f(t+r) - f(t)|}{r} \, dt$$

Hence, for every $x$,

$$\lim_{r \to 0} \frac{1}{r} \int_0^x[f(t+r) - f(t)] \, dt = 0$$

Since for almost every $x$,

$$F'(x) = \lim_{r \to 0} \frac{F(x+r) - F(x)}{r}$$

it follows that the limit of the first integral on the RHS of (*) must exist, and for almost every $x$,

$$f(x) = F'(x) = \lim_{r \to 0} \frac{1}{r} \int_0^r f(t) \, dt$$

Therefore, $F$ is almost everywhere a linear function and $f$ is almost everywhere constant.