Median–mean gap for convex Lipschitz concentration

The Lipschitz inequality makes $f$ continuous on the cube and gives $\mathrm{\mid }f(x)\mathrm{\mid }\le \mathrm{\mid }f(0)\mathrm{\mid }+\sqrt{d}$ there. Coordinatewise clipping $c(u{)}_i=\min \{1,\max \{0,u_i\}\}$ is continuous and takes ${\mathbb{R}}^d$ into the cube, so $f\circ c$ is continuous on ${\mathbb{R}}^d$. By [F3] it is Borel measurable. The vector $X$ is measurable: inverse images of open boxes are finite intersections of coordinate inverse images, and every open subset of ${\mathbb{R}}^d$ is a countable union of rational open boxes. Therefore $f(X)=(f\circ c)(X)$ is measurable. Put $C=\mathrm{\mid }f(0)\mathrm{\mid }+\sqrt{d}$. The constant function $C{\chi }_{\mathrm{\Omega }}$ is nonnegative simple and measurable, since its inverse images of Borel sets are either $\mathrm{\Omega }$ or $\mathrm{\varnothing }$. By [F4], ${\int }_{\mathrm{\Omega }}C{\chi }_{\mathrm{\Omega }}d\mathbb{P}=C\mathbb{P}(\mathrm{\Omega })=C<\mathrm{\infty }$. Since $\mathrm{\mid }f(X)\mathrm{\mid }\le C{\chi }_{\mathrm{\Omega }}$, monotonicity in [F3] gives ${\int }_{\mathrm{\Omega }}\mathrm{\mid }f(X)\mathrm{\mid }d\mathbb{P}\le C<\mathrm{\infty }$. Thus [F3] establishes integrability.

Apply [F1] with the given median and then [F2] with $Z=f(X)$, $m=Mf(X)$ and $L=1$. Step 1.1 supplies integrability, so $\mathrm{\mid }\mathbb{E}f(X)-Mf(X)\mathrm{\mid }<4$, and hence the asserted weak inequality follows.