Return to Revisions

Since $f$ is increasing, $\mu(0,x)=f(x)$ defines a (here: continuous) measure on $[0,c]$, which we can also view as a measure on $\mathbb R$ with support in this set. By Fubini, the RHS equals $$ \frac{1}{t}\int_0^1 dx\, f(x)\int_x^{x+t} d\mu(s)=\frac{1}{t}\int_0^{1+t}d\mu(s)\int^s_{s-t} dx\, f(x) . $$ Since $f$ is increasing and bounded, $(1/t)\int_{s-t}^s f(x)\, dx$ increases to a limit as $t\to 0+$ (at Lebesgue almost every $s$, the limit equals $f(s)$, by Lebesgue's differentiation theorem, but this doesn't matter here). Hence monotone convergence shows that $$ \int_0^1d\mu(s)\frac{1}{t} \int^s_{s-t} dx\, f(x) $$ converges. Finally, $(1/t)\int_1^{1+t} \ldots \to 0$ since $\mu(1,1+t)\to 0$ and $f$ is bounded.