\phi_t\,d\mu^\varphi$ for suitable test functions $v,w$.)

Recall that $\tilde Y=Y\times[0,1]$. We define the probability measure $\tilde\mu=\mu\times{\rm Lebesgue}$ on $\tilde Y$. Note that in the infinite measure case $\mu^\varphi|_{\tilde Y}=\tilde\mu$, whereas in the finite measure case $\mu^\varphi|_{\tilde Y}=(1/{\bar\varphi})\tilde\mu$.

Define $\tilde F:\tilde Y\to\tilde Y$ by setting $\tilde F(y,u)=(Fy,u)$. Note that $\tilde F(y,u)=f_{\varphi(y)}(y,u)$. Define $\tilde\varphi:\tilde Y\to{{\mathbb R}}^+$, $\tilde\varphi(y,u)=\varphi(y)$. Then $\tilde F=f_{\tilde\varphi}$. Let $\tilde R$ denote the transfer operator corresponding to the map $\tilde F:\tilde Y\to\tilde Y$ ($\int_{\tilde Y}\tilde  Rv\,w\,d\tilde \mu=\int_{\tilde Y} v\,w\circ \tilde F\,d\tilde \mu$). Given $v\in L^1(\tilde Y)$ and $u\in [0,1]$ we define $v^u\in L^1(Y)$, $v^u(y)=v(y,u)$. It is easily verified that $$\begin{aligned}
 \label{eq-Rtilde}
(\tilde R v)(y,u)=(Rv^u)(y).\end{aligned}$$

#### Renewal operators

For $t>0$, define $T_t,U_t:L^1(\tilde Y)\to L^1(\tilde Y)$ by setting $$T_tv = 1_{\tilde Y}L_t(1_{\tilde Y}v), \quad
U_tv = 1_{\tilde Y}L_t(1_{\{\tilde\varphi>t\}}v).$$ For $s\in{{\mathbb C}}$, define the families of operators on $L^1(\tilde Y)$, $$\hat R(s)v=\tilde R(e^{-s\tilde\varphi}v), \quad
\hat T(s)v=\int_0^\infty e^{-st}T_tv\,dt, \quad
\hat U(s)v=\int_0^\infty e^{-st}U_tv\,dt.$$ Note that $\hat R$, $\hat T$, $\hat U$ are analytic on ${{\mathbb H}}=\{{\operatorname{Re}}s>0\}$ and that $\hat R$ is well-defined on $\overline{{\mathbb H}}=\{{\operatorname{Re}}s\ge0\}$. We also define $\hat R_0(s):L^1(Y)\to L^1(Y)$ for $s\in\overline{{\mathbb H}}$: $\hat R_0(s)v=R(e^{-s\varphi}v)$.

\[prop-renew\] $\hat\rho(s)=\int_{\tilde Y}\hat T(s)v\, w\,d\mu^\varphi$ for $v\in L^1(\tilde Y)$, $w\in L^\infty(\tilde Y)$, $s\in{{\mathbb H}}$.

We have $\rho(t)=\int_{\tilde Y}v\,w\circ f_t\,d\mu^\varphi
