The continuous survival and cumulative-hazard identity

Since $T\ge 0$ and $S_T(0)=1$, $P_T(T=0)=0$. The density law gives $P_T(T\le x)={\int }_{[0,x]}f(u)du$; [F8] makes both endpoint integrals zero, so this equals ${\int }_0^{x}f(u)du$. Therefore $S_T(x)=1-{\int }_0^{x}f(u)du$ for every $x\ge 0$. This integral is at most $1$, so $f\in L^1[0,x]$ for each $x>0$; [F3] then makes $S_T$ absolutely continuous on each compact interval.

The identity is immediate at $t=0$, so fix $0<t<{\tau }_{\ast }$. The survival function is nonincreasing and positive on $[0,t]$, so $S_T(u)\ge S_T(t)>0$ there. For every $a\in \mathbb{R}$, the set $\{u\in [0,t]:S_T(u)>a\}$ is downward closed in $[0,t]$, hence is empty, all of $[0,t]$, or an interval with one endpoint at $0$; each is Lebesgue measurable by [F9]. Since the upper rays generate the Borel sigma-algebra, $S_T$ is Lebesgue measurable on $[0,t]$. The reciprocal map is continuous on $[S_T(t),1]$, so its inverse images of Borel sets are Borel; the preimage identity then makes $1\mathrm{/}{S}_T$ measurable. [F10] gives measurability of the product $f\mathrm{/}{S}_T$. Consequently ${\int }_0^t\frac{f(u)}{S_T(u)}du\le \frac{1}{S_T(t)}{\int }_0^{t}f(u)du<\mathrm{\infty }\mathrm{.}$ Thus ${\mathrm{\Lambda }}_T(x):={\int }_0^{x}f(u)\mathrm{/}{S}_T(u)du$ is absolutely continuous on $[0,t]$ by [F3].

By [F2] and step 1.1, $S_T^{\mathrm{\prime }}=-f$ almost everywhere on $[0,t]$. For $x,y\in [S_T(t),1]$, [F4] gives $\mathrm{\mid }\log x-\log y\mathrm{\mid }\le \frac{\mathrm{\mid }x-y\mathrm{\mid }}{S_T(t)}\mathrm{.}$ This Lipschitz bound and absolute continuity of $S_T$ show directly from the finite disjoint-interval definition that $\log \circ S_T$ is absolutely continuous. Applying [F2] to the integrable function $f\mathrm{/}{S}_T$ gives ${\mathrm{\Lambda }}_T^{\mathrm{\prime }}=f\mathrm{/}{S}_T$ almost everywhere. At almost every point where $S_T^{\mathrm{\prime }}$ exists, [F5] yields $(\log S_T{)}^{\mathrm{\prime }}=S_T^{\mathrm{\prime }}\mathrm{/}{S}_T=-f\mathrm{/}{S}_T=-{\mathrm{\Lambda }}_T^{\mathrm{\prime }}$.

Applying [F6] to $\log \circ S_T$ on $[0,t]$ gives $\log S_T(t)-\log S_T(0)=-{\int }_0^t\frac{f(u)}{S_T(u)}du\mathrm{.}$ Since $S_T(0)=1$ and [F11] gives $\log 1=0$, exponentiating and using [F11] yields the identity for $0<t<{\tau }_{\ast }$; the case $t=0$ was handled in step 1.2.

If ${\tau }_{\ast }<\mathrm{\infty }$, continuity from [F7] makes the zero set $Z:=\{t\ge 0:S_T(t)=0\}$ closed. Since $Z$ is nonempty and bounded below, ${\tau }_{\ast }:=\inf Z$ belongs to $Z$: for every $n\in \mathbb{N}$ choose $z_n\in Z$ with ${\tau }_{\ast }\le z_n<{\tau }_{\ast }+1\mathrm{/}(n+1)$, so $z_n\to {\tau }_{\ast }$ and continuity gives $S_T({\tau }_{\ast })=0$. Nonincreasing survival [F1] then gives $S_T(t)=0$ for $t\ge {\tau }_{\ast }$. Step 3.1 gives ${\mathrm{\Lambda }}_T(t)=-\log S_T(t)$ for $t<{\tau }_{\ast }$. Since $S_T(t)\to 0$ by continuity at ${\tau }_{\ast }$ and [F4], this tends to $+\mathrm{\infty }$ as $t\uparrow {\tau }_{\ast }$. The stated extended convention therefore gives $S_T(t)=e^{-{\mathrm{\Lambda }}_T(t)}$ also for $t\ge {\tau }_{\ast }$.

The identity holds throughout the positive-survival interval and at a finite terminal endpoint under the stated extended convention.