This has to do with the definition of the transmission coefficient $T$ in quantum scattering experiment. What we want to know is how a particle will behave when it's "shot" at the barrier. In a general situation, part of the wave function probability will reflect off the barrier, and part will be transmitted. However, for your potential, as $t\xrightarrow{}\infty$, all of the probability will be reflected. Note that we are NOT talking about a static situation. The wave function changes with time, so there will be some points in time when there is a non-zero probability for the particle to be found in the region where $V(x)=V_0$. However, after a long time has past, all of the probability will be reflected.
If the barrier were not infinitely thick, there would be some probability of transmission. For example, a particle certainly could tunnel through the potential
$$V=\begin{cases}
0 & x<0 \\
V_0 & 0\leq x < a \\
0 & a\leq x.
\end{cases}$$
Therefore, part of the probability could escape toward large $x$.
Edit:
For those who want more information, I'll do an explicit calculation. $R$ and $T$ are defined in the following way. We take a particle, localized on one side of the barrier and sharply peaked in momentum space, and allow it to travel toward the barrier. We then define $T$ as the probability of finding the particle on the opposite side of the barrier after waiting an infinitely long time. $R$ is defined as the probability that the particle will be found on the original side.
A standard way to approach the problem of calculating $R$ and $T$ (which is covered in most introductory textbooks) is to consider a slightly different situation. We look at the stationary state solutions for this potential (the energy eigenstates), and use $R=\frac{j_R}{j_I}$ and $T=\frac{j_T}{j_I}$, where $j_I$ is the probability current of the part of the wave function that is incident on the barrier, $j_R$ is the current of the reflected wave function, and $j_T$ is the current of the transmitted wave function.
The solution in the region where $V=0$ contains both $e^{ikx}$ and $e^{-ikx}$ terms, which give probability currents moving both to the left and to the right. The solution in the region where $V=V_0$ is a falling exponential $\psi_T(x)=e^{-\alpha x}$. The associated probability current is proportional to
$$j_T\propto \psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}=\psi\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi}{\partial x}=0$$
because the wave function is purely real ($\psi_T=\psi^*_T$). If you add time dependence, $\psi_T$ will rotate in phase space and acquire an imaginary part, but this imaginary part has no dependence on $x$, so it passes through the derivatives and does not affect the identity above. Therefore, $T=0$.
