Absolute real powers are Borel measurable and convex

For $p>1$ and $u>0$, differentiation gives $h^{\mathrm{\prime }}(u)=p{u}^{p-1}$ and $h^{\mathrm{\prime }\mathrm{\prime }}(u)=p(p-1)u^{p-2}\ge 0$ for $h(u)=u^p$. Thus $h$ is nondecreasing and convex on $(0,\mathrm{\infty })$. The derivative and second-derivative hypotheses hold at every positive $u$.

For $0<u\le 1$, $\log u\le 0$, hence $p\log u\le \log u$ and $0<u^p=\exp (p\log u)\le \exp (\log u)=u$. The last equality is the inverse identity [F16]. The squeeze theorem gives $u^p\to 0$ as $u\downarrow 0$. With $h(0)=0$, this extends $h$ continuously to $[0,\mathrm{\infty })$.

For $a,b\ge 0$ and $0\le \lambda \le 1$, apply positive-half-line convexity to $a+\epsilon ,b+\epsilon$ and let $\epsilon \downarrow 0$ using step 1.2: $h(\lambda a+(1-\lambda )b)\le \lambda h(a)+(1-\lambda )h(b)$. The inequality remains valid at weights zero and one, where it is equality. Monotonicity extends to zero because $h\ge 0=h(0)$.

For real $x,y$, [F10]–[F11] give $\mathrm{\mid }\lambda x+(1-\lambda )y\mathrm{\mid }\le \lambda \mathrm{\mid }x\mathrm{\mid }+(1-\lambda )\mathrm{\mid }y\mathrm{\mid }$. Apply monotonicity and then step 2.1 to get $\mathrm{\mid }\lambda x+(1-\lambda )y{\mathrm{\mid }}^p\le \lambda \mathrm{\mid }x{\mathrm{\mid }}^p+(1-\lambda )\mathrm{\mid }y{\mathrm{\mid }}^p$. For $p=1$ the same inequality is already precisely the triangle inequality with the scalar absolute values evaluated. Thus [F12] proves convexity for every $p\ge 1$.

The function is finite by [F1]. Continuity of absolute value [F13] and continuity of $h$ (steps 1.1–1.2, or the identity for $p=1$) imply continuity of $h(\mathrm{\mid }t\mathrm{\mid })$: choose an output tolerance for $h$ at $\mathrm{\mid }t\mathrm{\mid }$ and then the corresponding input tolerance for absolute value. Consequently all Borel preimages are Borel by [F14], which is exactly [F15].