The Geroch volume function: a continuous strictly increasing time function whose levels are Cauchy sets

(Well-definedness.) Let $p\in M$. The set $I^{-}(p)$ is open and nonempty (it contains the points $x_{\epsilon }^{-}\ll p$ of [L4]), so $0<\mu (I^{-}(p))$ by positivity of the measure [L2]; and $\mu (I^{-}(p))<1$: the point $x_{\epsilon }^{+}\gg p$ of [L4] does not lie in $J^{-}(p)$, because $x_{\epsilon }^{+}\le p$ together with $p\ll x_{\epsilon }^{+}$ would produce a nonconstant closed causal curve through $p$, contradicting [L1]; since $J^{-}(p)$ is closed by [L3] and $x_{\epsilon }^{+}\mathrm{\notin }{J}^{-}(p)$, a whole neighbourhood of $x_{\epsilon }^{+}$ is disjoint from $I^{-}(p)$, so $\mu (I^{-}(p))\le 1-\mu (\text{that neighbourhood})<1$. The same holds for $I^{+}(p)$. Hence $0<\mu (I^{\pm }(p))<1$ and $t(p)$ is a well-defined real number.

(Future decay of the future volumes.) Let $\gamma :I\to M$ be future-inextendible and fix $s_0\in I$. The family $\{I^{+}(\gamma (s)){\}}_{s\ge s_0}$ is decreasing, and each term has measure at most one. Suppose ${\inf }_{s\ge s_0}\mu (I^{+}(\gamma (s)))\ge \epsilon >0$ for some $\epsilon$. Choose an increasing cofinal sequence $s_j$ in the future parameter tail (a real interval admits one). The decreasing sequence $I^{+}(\gamma (s_j))$ has intersection $A:={\bigcap }_{s\ge s_0}{I}^{+}(\gamma (s))$ by cofinality and nesting. Measure continuity from above [L7] gives $\mu (A)\ge \epsilon$, so $A\mathrm{\ne }\mathrm{\varnothing }$; choose $z\in A$, so that $\gamma (s)\ll z$ for all $s\ge s_0$. Then $\gamma ([s_0,\sup I))\subseteq J^{+}(\gamma (s_0))\cap J^{-}(z)$, a compact causal diamond, so by [L5] the $h$-length of every subcurve of $\gamma$ is at most $L(K)<\mathrm{\infty }$ with $K:=J^{+}(\gamma (s_0))\cap J^{-}(z)$. Put $F(s)={\mathrm{\ell }}_h(\gamma {\mathrm{\mid }}_{[s_0,s]})$ and $L_{\mathrm{\infty }}={\sup }_{s\ge s_0}F(s)\le L(K)$. Integral additivity gives ${\mathrm{\ell }}_h(\gamma {\mathrm{\mid }}_{[s,t]})=F(t)-F(s)\le L_{\mathrm{\infty }}-F(s)$ for $s<t$, which tends to zero as $s\to \sup I$. Hence [L6] makes the image points along an increasing cofinal parameter sequence Cauchy; they converge to $z_0$ by completeness. The same distance bound and cofinality show that the entire tail $\gamma (s)$ tends to $z_0$, contradicting future-inextendibility. No change of parametrization is needed. Therefore $\mu (I^{+}(\gamma (s)))\to 0$ as $s\to \sup I$.

(Continuity.) If $p_k\to p$ then $\mu (I^{\pm }(p_k))\to \mu (I^{\pm }(p))$ by [L2], and the logarithm is continuous on $(0,\mathrm{\infty })$ by [L7]; the difference of two continuous functions is continuous, so $t$ is continuous. This proves (a).

(Strict monotonicity.) Let $p\le q$ with $p\mathrm{\ne }q$ and choose a future causal curve $\gamma :[0,1]\to M$ with $\gamma (0)=p$, $\gamma (1)=q$. Fix $s_0\in (0,1)$ and put $x:=\gamma (s_0)$; then $p\le x\le q$, and $p\mathrm{\ne }x$, $x\mathrm{\ne }q$ because equality would give a nonconstant closed causal subcurve, contrary to [L1]. We show $\mu (I^{-}(q))>\mu (I^{-}(p))$ and $\mu (I^{+}(p))>\mu (I^{+}(q))$. For the first: $I^{-}(p)\subseteq I^{-}(q)$, since $y\ll p\le q$ gives $y\ll q$ by [L2]. If $I^{-}(q)$ were contained in $J^{-}(p)$, then for every point $x_{\epsilon }^{-}\ll x$ of [L4] we would have $x_{\epsilon }^{-}\ll x\le q$, hence $x_{\epsilon }^{-}\ll q$ and therefore $x_{\epsilon }^{-}\in I^{-}(q)\subseteq J^{-}(p)$, that is, $x_{\epsilon }^{-}\le p$; letting $\epsilon \downarrow 0$ and using closedness of $\le$ [L3] gives $x\le p$, which with $p\le x$ yields a nonconstant closed causal curve through $x$ (nonconstant because $p\mathrm{\ne }x$), contradicting [L1]. Hence $I^{-}(q)\setminus J^{-}(p)$ is nonempty; it is open (as the intersection of the open $I^{-}(q)$ with the complement of the closed $J^{-}(p)$) and is contained in $I^{-}(q)\setminus I^{-}(p)$, so it has positive $\mu$-measure, and additivity gives $\mu (I^{-}(q))\ge \mu (I^{-}(p))+\mu (I^{-}(q)\setminus J^{-}(p))>\mu (I^{-}(p))$. For the second: $I^{+}(q)\subseteq I^{+}(p)$ by the dual push-up, and if $I^{+}(p)\subseteq J^{+}(q)$ then every $x_{\epsilon }^{+}\gg x$ satisfies $x_{\epsilon }^{+}\gg x\ge p$, hence $x_{\epsilon }^{+}\gg p$ and $x_{\epsilon }^{+}\in I^{+}(p)\subseteq J^{+}(q)$, that is $x_{\epsilon }^{+}\ge q$; closedness of $\le$ gives $x\ge q$, which with $x\le q$ is again a closed causal curve, contradicting [L1]. So $I^{+}(p)\setminus J^{+}(q)$ is a nonempty open subset of $I^{+}(p)\setminus I^{+}(q)$ and $\mu (I^{+}(p))>\mu (I^{+}(q))$. Since $L$ is strictly increasing, $t(p)=L(\mu (I^{-}(p)))-L(\mu (I^{+}(p)))<L(\mu (I^{-}(q)))-L(\mu (I^{+}(q)))=t(q)$. This proves (b).

(Future growth of the past volumes and the limit at the future end.) With $\gamma$, $s_0$ as in step 1.2: for $s\ge s_0$ one has $I^{-}(\gamma (s_0))\subseteq I^{-}(\gamma (s))$ because $y\ll \gamma (s_0)\le \gamma (s)$ gives $y\ll \gamma (s)$ by [L2], hence $\mu (I^{-}(\gamma (s)))\ge c:=\mu (I^{-}(\gamma (s_0)))>0$ by monotonicity and step 1.1. Combining with step 1.2, $t(\gamma (s))=L(\mu (I^{-}(\gamma (s))))-L(\mu (I^{+}(\gamma (s))))\ge L(c)-L(\mu (I^{+}(\gamma (s))))\mathrm{.}$ Since $\mu (I^{+}(\gamma (s)))\to 0+$ and the logarithm tends to $-\mathrm{\infty }$ at $0+$ [L7], the second term satisfies $-L(\mu (I^{+}(\gamma (s))))\to +\mathrm{\infty }$, while $L(\mu (I^{-}(\gamma (s))))\ge L(c)$ remains bounded below; hence $t(\gamma (s))\to +\mathrm{\infty }$ as $s\to \sup I$.

(The limit at the past end.) Apply steps 1.2 and 2.3 in the time-reversed orientation: the time reverse of $\gamma$ is future-inextendible, its future volumes are the $I^{-}$-volumes of $\gamma$, and its past volumes the $I^{+}$-volumes; hence $\mu (I^{-}(\gamma (s)))\to 0$ as $s\to \inf I$ and, with $\mu (I^{+}(\gamma (s)))\ge \mu (I^{+}(\gamma (s_1)))>0$ for $s\le s_1$, one gets $t(\gamma (s))\to -\mathrm{\infty }$ as $s\to \inf I$. This proves (c) for future-directed curves; a past-directed inextendible causal curve becomes future-directed after reversing its parameter, and the limits are interchanged accordingly.

(Level sets are Cauchy.) Let $c\in \mathbb{R}$ and let $\gamma :I\to M$ be inextendible and causal; reversing the parameter if necessary, assume it is future-directed. By step 2.3 and step 3.1 the continuous function $s\mapsto t(\gamma (s))$ on the interval $I$ tends to $+\mathrm{\infty }$ at $\sup I$ and to $-\mathrm{\infty }$ at $\inf I$; since the intermediate value theorem [L7] applies to continuous real functions on connected domains, for every $c$ between the two limits there is $s\in I$ with $t(\gamma (s))=c$. Hence every level set is met at least once. It is met at most once: if $s_1<s_2$ with $t(\gamma (s_1))=t(\gamma (s_2))=c$, then $\gamma (s_1)\le \gamma (s_2)$ with $\gamma (s_1)\mathrm{\ne }\gamma (s_2)$ (equality would be a nonconstant closed causal curve on $[s_1,s_2]$, contradicting [L1]), so strict monotonicity, step 2.2, gives $t(\gamma (s_1))<t(\gamma (s_2))$, a contradiction. Therefore every inextendible causal curve meets $t^{-1}(c)$ in exactly one parameter value, so $t^{-1}(c)$ is a Cauchy set; acausality is the special case of two points $y\le y^{\mathrm{\prime }}$ of the level set with $y\mathrm{\ne }{y}^{\mathrm{\prime }}$, which strict monotonicity forbids. This proves (d).