Temporal functions are time functions, and time functions forbid closed causal curves

(Pairing of the gradient with future causal vectors.) Put $w:=\mathrm{\nabla }u$. By [L3], $g(w,w)<0$ and $g(w,N)>0$, so $w$ is past-directed timelike and $n:=-w\mathrm{/}\sqrt{-g(w,w)}$ satisfies $g(n,n)=-1$ and $g(n,N)=-g(w,N)\mathrm{/}\sqrt{-g(w,w)}<0$; that is, $n$ is an observer. Let $v$ be any nonzero future causal vector; by [L1], $-g(n,v)>0$, that is $g(w,v)\mathrm{/}\sqrt{-g(w,w)}>0$, hence $g(w,v)>0$. By [L2] this is $du(v)>0$, which proves (a).

(The composite and its derivative.) Apply [L4] with the scalar target $P=\mathbb{R}$ and smooth map $F=u$. The composite $u\circ \gamma$ is locally absolutely continuous and $(u\circ \gamma {)}^{\mathrm{\prime }}(t)=du({\gamma }^{\mathrm{\prime }}(t))$ at almost every $t$. At every such $t$ at which in addition ${\gamma }^{\mathrm{\prime }}(t)$ is nonzero and future causal, (a) gives $du({\gamma }^{\mathrm{\prime }}(t))>0$; by [L4] the set of such $t$ has full measure, so $(u\circ \gamma {)}^{\mathrm{\prime }}(t)>0$ almost everywhere, and in particular the derivative is everywhere nonnegative almost everywhere.

(Strict monotonicity.) Let $a<b$ and suppose that $\gamma$ is not constant on $[a,b]$; restrict $\gamma$ to $[a,b]$, a future causal curve. Put $F:=u\circ \gamma {\mathrm{\mid }}_{[a,b]}$, which is absolutely continuous by step 2.1. By [L5], $F(b)-F(a)={\int }_a^b{F}^{\mathrm{\prime }}(t)dt\ge 0$. Suppose for contradiction that $F(b)=F(a)$. By [L5] and step 2.1, $0=F(b)-F(a)={\int }_a^b{F}^{\mathrm{\prime }}(t)dt$ with $F^{\mathrm{\prime }}\ge 0$ almost everywhere, so [L6] gives $F^{\mathrm{\prime }}=0$ almost everywhere on $[a,b]$. On the other hand, the set of $t\in (a,b)$ at which ${\gamma }^{\mathrm{\prime }}(t)$ is nonzero future causal has full measure in $[a,b]$, and at each such point $F^{\mathrm{\prime }}(t)=du({\gamma }^{\mathrm{\prime }}(t))>0$ by (a), so $F^{\mathrm{\prime }}>0$ on a nonempty full-measure set, contradicting $F^{\mathrm{\prime }}=0$ almost everywhere. Hence $F(b)>F(a)$, which is the inequality of (b).

(Time function, achronality and causality.) By step 3.1, $u$ is a time function in the sense of clause (d) of Chronology, causality, strong causality, global hyperbolicity and time functions. If $p\ll q$, choose a future timelike curve from $p$ to $q$; it is not constant on its parameter interval, because on a nondegenerate interval it has nonzero future timelike velocity at almost every parameter value (Causal curves, chronological and causal reachability, and inextendibility clause (b)), so step 3.1 gives $u(p)<u(q)$. Hence two distinct points of a level set $u^{-1}(c)$ are never chronologically related in either direction, that is, every level set is achronal. If $\gamma :[a,b]\to M$ is a nonconstant future causal curve with $\gamma (a)=\gamma (b)$, then $a<b$ and step 3.1 gives $u(\gamma (a))<u(\gamma (b))=u(\gamma (a))$, a contradiction; hence no such curve exists, and $(M,g,N)$ is causal and chronological by clause (a) of Chronology, causality, strong causality, global hyperbolicity and time functions. This proves (c) and completes the proof.