Weak topologies and weak convergence - Looking for feedbacks I am currently trying to get exactly what the weak and the weak* topologies are, in particular in connection to the concept of weak convergence in measure, however I am not completely sure on what I have got so far. So here, there is a summary of what I understood. I disseminated the text with some numbered questions, to point out what I am not sure about this summary.
Any feedback will be most welcome!
Edit: The answers I received have been quite useful and stimulating (e.g. I ended up looking at the Kelley-Namioka text), but I would love to get something more specific. Hence, I am adding a bounty. Thus, I slightly changed the text, because I actually changed my "opinion" regarding some issues.  
Edit 2: I know that editing put a question again on the top, however I did not edit with this purpose. I actually noticed that I did not get any new answer, and, considering I actually changed my view on some issues (I am studying the topic intensively these days), I changed the text improving (IMO) the content, without affecting any new answer. Moreover, I add some straightforward questions that arose out of these hours of study, in order to make clear what I am looking for, and I put the rest of the text as block, for those who want to see what I ended up with.
Questions: 
1. Isn’t it problematic to use the inner product notation when we deal with dual pairs and weak topologices, considering that we work in this case mainly with Banach spaces, and not all norm norms are associated to an inner product?  
2. Does it ever enter in the discussion on weak topologies the concept of algebraic dual?  
3. If by definition $X^*$ is the topological dual of $X$, hence it is the set of all continuous functionals on $X$, isn’t by definition that each $x^* \in X^*$ is actually continuous? Why do we need the weak topology on $X$?
[The same problem does not really apply to the weak* topology on $X^*$ that makes each evaluation functional $e_x: X \to \mathbb{R}$ continuous, because those evaluation functionals are not continuous by definition.]  
4. I see in which way the fact that the bidual $X^{**}$ can be way bigger than $X$ when we deal with weak topologies can be relevant, namely if $X = X^{**}$ then weak and weak* topologies coincide. However, I don’t see how can be that $X \subset X^{**}$.  
5. Related to the previous problem is the fact that when we move from dual pairs and weak topologies to weak convergence in measure, I don’t see why the weak convergence in measure coincides with endowing the reference metric space $(X,d)$ with the weak* topology, and not simply the weak topology. Thus the question should be, how do we decide what is the main space, what is the dual one, and which one is the one endowed with a topology, in order to come up with the concept of weak convergence? Is it through the Riesz Representation theorem?
(...and now I should end up looking exactly what that representation thoerem does...)
6. I read that what is called weak convergence should be called weak* convergence, because it is based on the weak* topology. Why, actually?
[True, this question can be considered a repetition of the previous] 
In the following, there are my thoughts on weak topologies and weak convergence. I hope they are correct or interesting. I put them as a spoiler in order not to scare who would like to answer.


In general we have the following definition of initial (or weak) topology: given a set $X \neq \varnothing$, a family of topological spaces $\{ (Y_i, \tau_i ) \}_{i \in I}$, and for every $i \in I$ a function $f_i : X \to Y_i$, the initial (or weak) topology is the weakest topology on $X$ that makes all the functions $f_i$ continuous.
  Notice that, if $\mathcal{F} := \{ f | f: X \to \mathbb{R} \}$, then $X$ can be seen as set of real valued functions on $\mathcal{F}$, where each $x \in X$ is an evaluation functional $e_x : X^* \to \mathbb{R}$ with $e_x (f) = f(x)$. Thus, the weak topology on $\mathcal{F}$, denoted by $\sigma (\mathcal{F}, X)$, is identical to the relative topology on $\mathcal{F}$ as subset of $\mathbb{R}^X$ endowed with the product topology.  
Now, let $X$ a linear space with $x \in X$. Thus, let $X’$ be the algebraic dual of $X$, i.e. the vector space of all linear functionals on $X$, and let $X^*$ be the topological dual of $X$, i.e. $X^* := \{ x^* | x^*: X \to \mathbb{R}, x^* \text{ continuous} \}$ is, or – in plain english – the vector space of all continuous linear functionals on $X$, with $x^* \in X^*$.   
Notice that, for the following, for example Rudin points out that $\langle x , x^* \rangle \equiv x^*(x)$, thus we should have that $\langle x^* , x \rangle \equiv e_x (x^*)$. [a. Right?]
Thus, $X$ can be seen as a vector subspace of $\mathbb{R}^{X^*}$, in which case $\sigma (X, X^*)$ is the weak topology on $X$ (denoted by $w$), defined as
  $$ x_\alpha \overset{w}{\to} x \in X \Longleftrightarrow \forall x^* \in X^* , \langle x_a, x^* \rangle \to \langle x , x^* \rangle. $$
In the same vein, $X^*$ can be seen as a vector subspace of $\mathbb{R}^X$, in which case $\sigma (X^*, X)$ is the weak* topology on $X^*$ (denoted by $w^*$), defined as
  $$ x^{*}_\alpha \overset{w^*}{\to} x^* \in X \Longleftrightarrow \forall x \in X , \langle x^{*}_a, x \rangle \to \langle x^* , x \rangle. $$ 
Notice that in both cases we are just declining in different ways the original definition of initial topology given at the beginning. What we are changing is simply the "reference" space that makes all functions continuous.  Indeed, in the case of the weak topology, we have that $(X, \sigma(X, X^*)$ is the topological space that makes all the $x^* \in X^*$ continuous. In the case of the weak* topology, we have that $(X^*, \sigma(X^*, X)$ is the topological space that makes all the evaluation functionals $e_x$ from $X$ to $\mathbb{R}$ continuous.



I know (after quite some bookreading) that what can be found in the following text as the old point of view is the correct way of looking at the problem, however I keep on feel that it is the new point of view the correct way. In particular, I find what I wrote in the old point of view misleading. Indeed, I implicitly state that we have as dual pair $\langle X, C(X) \rangle$ and every $\mu$ is on $X$, but this does not make sense. Indeed, we should have that the dual pair is $\langle \Delta (X), C(X) \rangle$ and every $\mu$ is on $\Delta (X)$. But then, if the dual pair is really $\langle \Delta (X), C(X) \rangle$, according to the definition of weak and weak* topology I wrote down, here we are actually dealing with the weak topology! Hence, the dual pair should be $\langle C(X), \Delta(X) \rangle$, but why?



Old point of view:
Thus, if we take all these definitions and we use them to get what the weak convergence in measures is, we actually have that weak convergence in measure is the same as convergence in the weak* topology. [b. Right?]
Indeed, given a metric space $(X,d)$ a sequence of measures $(\mu_n)$ converges weakly to $\mu$  in $X$ if and only if for every $\phi \in C(X)$ (where $C(X)$ is the space of all continuous functions on $X$),
  $$ \lim_{n \to \infty} \int_X \phi d\mu_n = \int_X \phi d\mu, $$ 
  and here the $\mu_n$ should make the same job of our $x^{*}_n$ in our definition of the weak* topology. [c. Right?].



New point of view:
Thus, if we take all these definitions and we use them to get what the weak convergence in measures is, we actually have that weak convergence in measure is the same as convergence in the weak topology. [d. Right?]
Indeed, given a metric space $(X,d)$ a sequence of measures $(\mu_n)$ converges weakly to $\mu$  in $\Delta (X)$, where $\Delta (X)$ denotes the set of all probability measures on $X$, if and only if for every $\phi \in C(X)$ (where $C(X)$ is the space of all continuous functions on $X$),
  $$ \lim_{n \to \infty} \int_X \phi d\mu_n = \int_X \phi d\mu. $$
  Thus, in this setting the dual pair of spaces is $\langle \Delta (X) , C (X) \rangle$ and indeed the definition of weak convergence in $\Delta (X)$ ends up to be equivalent to endow $\Delta (X)$ with the weak topology (and not the weak* topology!).
  [e. Right?].

Thank you for your time and for your help!
 A: I would say "Yes" to all your questions.
The so-called weak* topology is an example of a weak topology.
Weak convergence of measures can be considered a weak* topology.
To clarify a new example of convergence of this kind, you should master that $\sigma (\mathcal{F}, X)$ notation. (I think it goes back to Mackey, and was popularized by Bourbaki.)
A: Dual Pair
Two vector spaces with a bilinear form satisfying:
$$\langle\cdot,\cdot\rangle:X\times Y\to\mathbb{C}:\quad X^\perp=(0),\,Y_\perp=(0)$$
(Informally, this way one guarantees that it separates points.)
Weak Topology
The initial topology generated by the dual:
$$\sigma(X;Y):=\tau\left(\bigcup_{y\in Y}\varepsilon_y^{-1}(\mathcal{T}_\mathbb{C})\right)\quad(\varepsilon_y:=\langle\cdot,y\rangle)$$
(This way one guatantees that they become continuous.)
Example
Consider a Banach space with its continuous dual:
$$\langle\cdot,\cdot\rangle:E\times E'\to\mathbb{C}:\quad\langle x,f\rangle:=f(x)$$
By the initial topology weak convergence boils down to:
$$x_n\rightharpoonup x\iff f(x_n)\to f(x)\quad(f\in E')$$
(That is in terms of product topology convergence by components.)
Remark
The difference between the weak and weak* topology lie in the chosen dual:
$$\langle x,f\rangle:=f(x):\quad\sigma(E';E)$$
$$\langle F,f\rangle:=F(f):\quad\sigma(E';E'')$$
(The comparison becomes solid on embedding the Banach space into its double dual.)
