Rectangular potential barrier Take the usual rectangular potential barrier, that is:
$$V(x)=0 \: \text{if} \: x<0 \: \text{or}\: \: x>a$$
$$V(x)=V_0 \: \text{if} \: 0\leq x \leq a.$$
I've looked at several notes and books and in everyone of them the books supposes, in the third region ($x\geq a$), a solution to the time-independent Schroedinger equation of the form:
$$\psi= Ae^{ikx}.$$
Advocating that "we don't want any wave travelling to the right". Now, it isn't immediately clear to me that the wavefunction can indeed be interpreted as a wave, and that is even travelling. What is precisely the boundary condition that permits me to say that the constant in front of the second exponential is $0$? If the wavefunction were normalizable then the usual condition of $\psi \rightarrow 0$ as $|x|\rightarrow \infty$ would do. But what in this case?
 A: The idea is the following. You have three zones. One before the barrier, $x<0$, with $V(x)=0$ (zone 1), one inside (or above) the barrier $0\leq x\leq a$ with $V(x)=V_0>0$ (zone 2), and one after the barrier with $V(x)=0$ (zone 3). 
Let's say that a particle comes from the left, with energy $0<E<V_0$, so it will go "inside" the barrier. Then, you'll have plain wavefunctions in the zones 1 and 3, where the energy is greater than the potential, and real exponential functions in zone 2, something like
$$\psi _1(x) = Ae^{ik_1x} + Be^{-ik_1x}$$
$$\psi _2(x) = Ce^{k_2x} + De^{-k_2x}$$
$$\psi _3(x) = De^{ik_1x} + Ee^{-ik_1x}$$
This comes from the solution of the time-indepent Schrödinger equation with a constant potential. From your question I assume you know how to reach this point (comment me if you don't). Note that I've used the same $k_1$ in zones 1 and 3 because the potential is the same.
Now you have to apply boundary conditions. Of course, $\psi$ must be continuous and derivable, so we should apply that:


*

*Between zones 1 and 2, in $x=0$
$$\psi_1(0) = \psi_2(0)$$
$$\psi_1'(0) = \psi_2'(0)$$

*Between zones 2 and 3, in $x=a$
$$\psi_2(a) = \psi_3(a)$$
$$\psi_2'(a) = \psi_3'(a)$$
Well, you have 6 constants, but only 4 boundaries. Of course, the final goal is to write the final wavefunction $\psi(x)$ in terms of only ONE constant (let's say $A$). To achieve this, I need 5 boundary conditions. But I only have 4. What should I do?
In this case I can eliminate one constant on zone 1 or 3. Whatever constant I want. Do you want to eliminate $B$? Do it. Maybe $D$? Do it then. The problem is not going to change. Try it and convince yourself. But I would choose to eliminate $E$, the wave in zone 3 that comes from right to left. Why? Simply because I imagine a particle coming from left. This will reflect between zones 1 and 2, and also between zones 2 and 3. But in zone 3 I only have transmission: only a particle going from left to right. Although any constant can be eliminated, we eliminate the one we want according to our classical thought. Make that constant 0. Now your 4 boundaries are enough.
Once you have the expression of the final $\psi(x)$ in terms of a constant. This final function will have plain waves in the zone 1 and also in 3, so it won't be normalizable in the usual way:
$$\int _{-\infty}^{+\infty}|\psi(x)|^2 dx = 1$$
because sines and cosines are not square-integrable. If you want to think on normalization, you can think on some alternative ways of normalization, for example: instead of an infinite $X$ axis, you could have very big box where $\psi(L)=\psi(-L)=0$. Then your function would be normalized. Of course, you must think that $L$ is large enough so you can think on a barrier (and not on a barried inside an infinite well). Also you can take a look to "rigged Hilbert spaces" for discussion on this non square-integrable functions.
In any case, you should forget about a normalization with plain waves. The big box concept will do the trick. Instead of that, in this kind of problems you're interested in the reflection and transmission coefficients. They're based on probability currents and this things, but in this simple cases can be calculated as
$$T = \dfrac{|B|^2}{|A|^2}$$
$$R = T - 1$$
Since $B$ is the wave which go left in zone 1 (reflection). 
You finally calculate $R$  writting $B$ and $A$ in terms of the constant you've chosen before (if you've chosen $A$, it will be easy :D). That will eliminate all the constants and say you how many particles will be able to go through the barrier and how many will reflect -that is, in fact, the thing you can measure in a real experiment.
A: To put it another way: typically with this sort of setup we want there to be an incoming wave from the left, $\psi_0 = e^{-i \omega t + i k x},$ which is partially "reflected" into an outgoing wave to the left $\psi_\ell = r e^{- i \omega t - i k x},$ and partly "transmitted" to an outgoing wave to the right, $\psi_r = t e^{- i \omega t + i k x}.$
The condition that these be travelling waves comes from the implicit $e^{-i\omega t}$ which is prefixing all of them, as a traveling wave has the form $f(x - v t)$ where $v$ is the velocity of the wave. The fact that this is the same for all three of these waves means that the scattering is elastic and is not consuming any energy, as it comes from the $i \hbar \partial_t \psi$ part of the wavefunction, enforcing that great Planck/Einstein law, $E = \hbar \omega.$ The $-i$ perfectly cancels the $i$, leaving just $\hbar \omega \psi = +\frac{\hbar^2 k^2}{2m} \psi = \frac {p^2}{2m} \psi.$
The condition that no wave is coming in from infinity on the right really is just saying "that's not what we're studying": we want only the left-hand-side to have an incoming wave so that we can study how the barrier reflects and transmits that wave. You could then have the barrier transmitting/reflecting two incoming waves, but it would be a superposition of their individual transmissions/reflections and would not be very interesting, once you've worked out the one-incoming-wave solution.
