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Let $h:\mathbb{R}\to \mathbb{R}$ be continuous, strictly increasing, and satisfy Lusin condition N. Thendefines a measure on the Lebesgue measurable sets, and $\nu \ll \lambda$. Moreover,for $a\le b$.

A nondecreasing function never decreases when its argument increases. Flat intervals are allowed, unlike for a strictly increasing function. Nondecreasing allocation functions permit critical-value payments in single-parameter mechanisms.

First, equality of two $U_0$ values forces equality of the corresponding $V_0$ values: neither strict comparison holds for $U_0$, and the same is then true for $V_0$ in both directions. Thus $V_0=g\circ U_0$ for a well-defined strictly increasing function $g$ on $J=U_0(\mathcal{P})$. Since $\mathcal{P}$ is a convex set and $U_0,V_0$ are affine functions, $J$ is a real interval andIf $J$ contains $u<v$, let $a=[g(v)-g(u)]/(v-u)>0$ and $b=g(u)-au$. The mixture identity gives $g(w)=aw+b$ for $w\in [u,v]$. For $w>v$, express $v$ as a convex combination of $u$ and $w$ and solve the same identity for $g(w)$; for $w<u$, express $u$ as a mixture of $w$ and $v$. Hence the formula holds on all of $J$, not merely between the chosen anchors. We obtain positive affine uniqueness of affine preference representations:If $U_0$ is constant, the same comparisons force $V_0$ to be constant; choose $a=1$ and the appropriate $b$. If the probability-measure set is empty, the assertion is vacuous. These degenerate cases do not require distinct anchors.

Fix $a<x<b$. Injectivity excludes $f(x)=c$ or $d$. If $f(x)<c$, the intermediate value theorem on $[x,b]$ gives a point other than $a$ with value $c$, contradicting injective function. If $f(x)>d$, the same theorem on $[a,x]$ gives a point other than $b$ with value $d$. ThereforeTo prove strict increase, take $x<y$. If $x=a$, the displayed endpoint ordering already gives $f(x)<f(y)$. Otherwise $f(a)<f(y)$, and the same interior-point argument on the subinterval $[a,y]$ yields $f(a)<f(x)<f(y)$. Hence $f$ is a strictly increasing function. This is the interval order theorem for continuous injections; the assumed endpoint order rules out strict decrease.

If $f$ is a strictly increasing function and $x\ne y$, one of $x<y$ or $y<x$ holds. The corresponding strict inequality of function values shows that $f$ is injective. Conversely, if $f(x)<f(y)$, then $x=y$ is impossible and $x>y$ would imply $f(x)>f(y)$. ThusTogether with strict increase, this says that such a function preserves and reflects the ordering of points.

PutBecause $dR/ds=-n\cdot v$, differentiation givesA massive charge has the subluminal speed $|v|<c$, so $F^{\prime }(s)>0$. Hence $F$ is strictly increasing, and $F(s)=t$ has at most one root. This is the uniqueness of retarded time for a subluminal source.