The hazard rate is defined by

$\lambda(t)=\frac{f(t)}{1-F(t)} =\lim_{\Delta\downarrow0}\frac{\mathbb{P}\bigl(t\le T<t+\Delta\mid T\ge t\bigr)}{\Delta},$

and integrating $\lambda=-\frac{d}{dt}\log\bigl(1-F(t)\bigr)$ gives the survival function

$S(t)=1-F(t)=\exp\left\{-\int_0^{t}\lambda(u)\,du\right\}.$

(C) is true, and (D) false, for the same reason. $T$ is a genuine (finite) lifetime, so $S(t)\to0$ as $t\to\infty$ . By $(\dagger)$ that happens precisely when

$\int_0^{\infty}\lambda(u)\,du=\infty .$

Conversely, any non-negative function with divergent integral is the hazard rate of some absolutely continuous lifetime distribution, via $(\dagger)$ . So the cumulative hazard must diverge — it is emphatically not finite, which kills (D). (A finite total hazard would leave $S(\infty)>0$ : a defective distribution with positive probability of never failing.)

(A) and (B) confuse a rate with a probability. For a continuous lifetime $\mathbb{P}(T=t)=0$ at every $t$ , so neither can be a probability of failing at $t$ ; and $\lambda(t)$ is not even bounded by $1$ — it may exceed $1$ freely. Statement (B) is the loose textbook gloss, correct only once one divides by $\Delta$ and takes a limit, i.e. it describes a conditional failure rate per unit time, not a probability.