Suppose $1 \leq p <\infty$. Let $(T_t)_{t \geq 0}$ be a strongly continuous semigroup of multiplication operators on $L^p(0,1)$ defined by $T_t(f)=m_t \times f$ where the function $m_t \colon [0,1] \to \mathbb{R}$ is measurable.
How show that there exists a measurable function $g$ on $[0,1]$ such that $m_t=e^{tg}$ almost everywhere for all $t \geq 0 $?
I tried the following. For any $t,t' \geq 0$, the relation $T_tT_{t'}(1_{[0,1]})=T_{t+t'}(1_{[0,1]})$ gives $m_t(x)m_{t'}(x)=m_{t+t'}(x)$ for almost all $x$ and $m_0=1_{[0,1]}$. By the characterization of the exponential function (problem of the measurability of $t \mapsto m_t(x)$?), we obtain $m_t(x)=e^{tg(x)}$ for some $g(x)$. Now, why $g$ is measurable and why we can define $g$ on all $[0,1]$ ?