$
\newcommand{\bR}{\mathbb{R}}
\newcommand{\bE}{\mathbb{E}}
\newcommand{\bT}{\mathbb{T}}
\newcommand{\bP}{\mathbb{P}}
\newcommand{\bF}{\mathbb{F}}
\newcommand{\cF}{\mathcal{F}}
\newcommand{\eps}{\varepsilon}
\newcommand{\diff}{\mathop{}\!\mathrm{d}}
\newcommand{\qtext}[1]{\quad\text{#1}}
\newcommand{\qtextq}[1]{\quad\text{#1}\quad}
\DeclareMathOperator*{\trace}{tr}
$
We fix $T \in (0, \infty)$ and let $\bT$ be the interval $[0, T]$. We consider measurable functions
\begin{align}
b &: \bT \times \bR^d \to \bR^d, \\
\sigma &: \bT \times \bR^d \to \bR^d \otimes \bR^m.
\end{align}

We assume that $b$ $\sigma$ are regular enough. Let $(B_t, t \ge 0)$ be a $m$-dimensional Brownian motion and $\bF := (\cF_t, t \ge 0)$ an admissible filtration on a probability space $(\Omega, \mathcal A, \bP)$. For $(s, x) \in [0, T) \times \bR^d$, we consider the SDE
$$
\begin{cases}
\diff X^x_{s,t} & = b (t, X^x_{s, t}) \diff t + \sigma (t, X^x_{s, t}) \diff B_t
\qtextq{for} t \in [s, T] , \\
X^x_{s,s} & = x .
\end{cases}
$$

The differential operator $(L_t)_{t \in \bT}$ is defined for $f \in C^2 (\bR^d)$ and $x \in \bR^d$ by
\begin{align}
L_t f (x) := \langle b (t, x), \nabla f (x) \rangle + \frac{1}{2} \trace ( a (t, x) \nabla^2 f (x) ) .
\end{align} 

The semigroup $(P_{s, t})_{0 \le s < t \le T}$ is defined for $x \in {\bR}^d$ and $f \in C_b ({\bR}^d)$ by
$$
P_{s, t} f (x) := \bE [f(X^x_{s, t})] .
$$

We consider two statements:

1. It holds for $f, g \in C^2_c (\bR^d)$ that
$$
\int_{\bR^d} (L_t f) g \diff x = \int_{\bR^d} f (L_t g) \diff x .
$$

2. It holds for $f, g \in C_c (\bR^d)$ that
$$
\int_{\bR^d} (P_{s, t} f) g \diff x = \int_{\bR^d} f (P_{s, t} g) \diff x .
$$

>Is it true that (1) implies (2)?

Thank you for your elaboration.