Normalization of wave function meaning...? I just have one question.
I'm doing a problem where I'm told to normalize a wave function, which is split up into two regions, namely where $r \leq r_0$ and $r > r_0$. My question is, why am I doing this? I'm not using any of the math I get out of it later on. The only thing I do after it is applying the continuous condition. So what I was thinking was, that by normalizing my wave function, I also showed that my wave function is in fact continuous. So is that the case, or am I just mistaken all together? 
 A: I think what you are asking whether the relationship 
$$ \mathrm{normalizable} \iff \mathrm{continuous}$$ 
holds, which is utterly wrong! The wave function has to be continuous*. Notwithstanding take $\psi(x)=H(x-1/2)-H(x+1/2)$, where $H(x)$ is the Heaviside step function.
$$ \implies \int_{-\infty}^\infty \mathrm{d}x \,\, ||{\psi(x)}||^2 = 1 $$ 
(Area of a square with sides 1)
Thus, although the function isn't continuous, it is normalizable.
Edit:
*As ACuriousMind points out the wave function, in general, need not be continuous, although in the physical world it has to be so.
A: 
My question is, why am I doing this?

Becase, by convention, we want the probability account for all possible outcomes to sum to unity. The fact that we will get some outcome at a measurement, is guaranteed, and with this normalization it reflects a total probability of 1.
A: Like gonenc pointed out your assumption that normalizing your wave function does not imply continuity. And yes you'll probably won't need the normalization factor in your further calculations.
The reason for you doing this could be consistency with the Interpretation of the wave function squared as a probability amplitude:
$$
P = \int|\psi(r,t)|^2 dr
$$
