Show the following addition to Lemma 19.16:\[\tau(t-)=\inf \{s \geqslant 0: a(s) \geqslant t\} \quad \text { and } \quad a(t-)=\inf \{s \geqslant 0: \tau(s) \geqslant t\}\]

and \(\tau(s) \geqslant t \Longleftrightarrow a(t-) \leqslant s\).

Data From Lemma 19.16

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Ex. 19.11 19.16 Lemma. Let T be the generalized inverse of a right continuous increasing function a:[0,00) [0, co]. a) T:[0,00) [0,oo] is increasing and right continuous; b) a(r(s))>s; c) a(t)st(S-) t}; e) a(t(s)) = suplus: T(u) = T(S)); f) T(a(t)) sup{w>t: a(w)= a(t)}; g) if a is continuous, a(t(s)) =s for all sea([0, o)). Proof. Before we begin with the formal proof, it is helpful to draw a picture of the situation, cf. Figure 19.1. All assertions are clear if t is a point of strict increase for the function a. Moreover, if a is flat, the inverser makes a jump, and vice versa. a) It is obvious from the definition that r is increasing. Note that ta(t)>s) Ut a(t)>s+c). = Therefore, inf{t > 0 a(t) > s) info infit > 0 a(t) > s + ) proving right continuity. b) Since a(u)>s for all u > r(s), we see that a(r(s)) > s. c) By the very definition of the generalized inverse T we have a(t)sa(t)>s-c vc>0 =>T(S-E) t VE> 0 T(S-)