Does bra-ket always assume all space? One thing I never understood about the bracket notation is the limits of the inner product. Given $ \langle \psi∣\psi \rangle$, what can I say about the limits of integration of the inner product?
Does $ \langle \psi∣\psi \rangle$ always assume all space i.e. $ \langle \psi∣\psi \rangle = 1$?
How can I specify/denote the region of integration if I am only interested the probability of a certain range $[a,b]$?
 A: I will try to answer this in the more general case where the configuration space is $\mathbb R^n$. In this case the Hilbert space of the quantum theory is $L^2(\mathbb R^n)$ with Lebesgue measure, and the inner product has the representation
$$(f,g) = \int_{\mathbb R^n}\overline{f(x)}g(x)\ \text d\lambda(x)$$
where $\lambda$ is the Lebesgue measure.
Let $U$ be any (Borel) subset of $\mathbb R^n$ (read any region of the configuration space $\mathbb R^n$), then the integral
$$\int_U\overline{f(x)}g(x)\ \text d\lambda(x)$$
can be rewritten using the characteristic function $\chi_U$ of $U$ as
$$\int_{\mathbb R^n}\overline{f(x)}\chi_U(x)g(x)\ \text d\lambda(x).$$
To $\chi_U$ one can associate a bounded operator $E_U$ that acts on the elements of the Hibert space as
$$(E_Ug)(x) = \chi_U(x)g(x),\qquad\forall g\in L^2(\mathbb R^n)$$
so that the sought integral takes the form
$$(f,E_Ug) = \int_U\overline{f(x)}g(x)\ \text d\lambda(x).$$
In Dirac notation this means that there exists a projection (i.e. self-adjoint and idempotent element of the algebra of (generalised) observables) such that the integral restricted to a certain region is
$$\langle\psi|E_U|\psi\rangle.$$
When $n=1$ and $U=[a,b]$, then one has to take the characteristic function of this interval, which is just the projection onto this portion of the whole real line.
A: 
How can I specify/denote the region of integration if I am only
  interested the probability of a certain range [a,b]?

If a particle is in a state $|\psi\rangle$, the probability of finding the particle in state $|\phi\rangle$ is given by (assuming the states are normalized)
$$P(|\phi\rangle) = |\langle \phi | \psi\rangle|^2$$
For example, the probability of finding the particle in a (1-D) position eigenstate is
$$P(x = x_0) =  |\langle x_0 | \psi\rangle|^2$$
Then, if you wish to find the probability of finding the particle between $a$ and $b$, you would 'sum' the probabilities of finding the particle in any of the position eigenstates in-between
$$P(a \le x \le b) = \int_a^b dx\;|\langle x | \psi\rangle|^2$$
But
$$\psi(x) = \langle x | \psi\rangle $$
thus
$$P(a \le x \le b) = \int_a^b dx\;|\psi(x)|^2$$
A: Ok, first, the inner product $\langle\psi|\psi\rangle=1$ tells you two things:
1. States in QM correspond to normalized vectors.
2. The overall phase is irrelevant.
and yes, if you calculate an inner product and deal with continuous variables, you should integrate over the whole space you observe. Otherwise your normalization doesn't make any sense, right?   
Second, what you actually trying to ask here is called a probability interpretation and for that you need a density operator, which is usually denoted as $\hat\rho = |\psi\rangle\langle\psi|$. Then you can calculate probability to find your object in some sate $|\phi\rangle$, i.e. you perform a protective measurement $|\phi\rangle\langle\phi|$, which leads you to the desired probability:
$$
P = {Tr}(\hat\rho|\phi\rangle\langle\phi|) = \langle\phi|\hat\rho|\phi\rangle = \langle\phi|\psi\rangle\langle\psi|\phi\rangle = |\langle\psi|\phi\rangle|^2
$$
Then comes a tricky part, when you start speaking about your object's state with defined coordinates, which people denote as $|x_0\rangle$, and say that you have a corresponding probability to find the object at this spot, while actually they mean "somewhere around the spot". In practice you cannot measure anything precisely, can you? In theory it could be said that you have a probability for your object to be found in some $\delta x$ neighborhood of $x_0$. (If someone knows a better way to define this, I will be glad to be taught.) Finally, after unnecessary explanations comes a real answer:    
What you have to integrate is not an inner product $\langle\psi|\psi\rangle$, but the probability density function, which is left after you wrote $\langle x|\psi\rangle\langle\psi|x\rangle$. Then you can integrate over any interval (or box) you want, and you will get something like an approximation of probability to  find your object in this region.
I think that the problem is hiding in the similarity of two integrals, when instead of vectors people write $\psi(x) = \langle x|\psi\rangle$, and
$$
\langle\psi|\psi\rangle = \int\limits_{-\infty}^\infty \langle \psi|x\rangle\langle x|\psi\rangle\, dx = \int\limits_{-\infty}^\infty |\psi(x)|^2 dx = 1, 
$$
$$
P([a,\, b]) = \int\limits_a^b \langle x|\psi\rangle\langle\psi|x\rangle\, dx = \int\limits_a^b |\psi(x)|^2 dx =\, ?, 
$$
and these are two different things, actually.  
Hope this will help.
P.S. Limits $(-\infty, \infty)$ are not general, they just mean that you look at the whole space.             
A: Bra ket notation has nothing to do with integrals.
If I have two vectors in 3D space, $\vec{v}$  and $\vec{w}$ , I can write them as $|v\rangle$ and $|w\rangle$ if I want.
In that case, I would write their dot product (a.k.a. inner product) as $\langle v | w \rangle$.
This dot product has nothing to do with integration.
When your vectors are functions then the dot product is an integral and it's usually the case that this integral goes from $-\infty$ to $\infty$.
If that's the case, then you can't represent integration over a different range with bra-ket.
At least, that's not what the notation means.
Here's a really nice way to think about this business:
The simple kinds of vectors you're used to thinking about are represented as lists of numbers.
For example, we can have an n-dimensional vector $|v\rangle$ which is represented as
$$|v\rangle \rightarrow \left( \begin{array}{c} v_1 \\ v_2 \\ \vdots \\ v_n \end{array} \right) \, .$$
Now suppose we have another vector $|w\rangle$ represented as
$$|w\rangle \rightarrow \left( \begin{array}{c} w_1 \\ w_2 \\ \vdots \\ w_n \end{array} \right) \, .$$
The dot product of these vectors is
$$\langle w | v \rangle = \sum_i v_i w_i^* \, .$$
Ok now suppose I have a vector $|f\rangle$ which is infinite dimensional and in fact has as many components as there are real numbers.
In other words, for each real number I have a component for $|f\rangle$.
Since I'm associating a value (the component of $|f\rangle$) to each real number, I might as well think of $|f\rangle$ as a function where $f(x)$ is the $x^{\text{th}}$ component of the vector $|f\rangle$.
From that point of view, if I have another vector $|g\rangle$, then the dot product between $|f\rangle$ and $|g\rangle$ is
$$\langle f | g \rangle = \int f^*(x) g(x) \, dx \, .$$
This integral is taken over the whole domain (usually $\mathbb{R}$ or $\mathbb{R}^n$) and it constitutes an $L^2$ inner product (technically Hermitian). 
There are lots of nice ways to understand all this business.
As you learn more you'll find out that different functions can be representations of the same vector in different bases, just like two different columns of numbers can be representations of regular vectors in different bases.
You'll also learn that the Fourier transform is a really common change of basis between sinusoids and delta functions.
