Intuition: Portmanteau-Theorem An important theorem in probability theory about weak convergence of measures is the Portmanteau-Theorem.
Why should it be true - intuitively - though?
EDIT: Our version of Portmanteau's Theorem is: The following statements are equivalent


*

*$\mu_n {\to} \mu$ weakly

*$\int f \ d \mu_n \to \int f \ d\mu$ for all uniformly continuous and bounded $f:S \to \mathbb{R}$

*$\limsup_{n \to \infty} \mu_n(F) \leq \mu(F)$ for all measurable closed subsets

*$\liminf_{n \to \infty} \mu_n(U) \geq \mu(U)$ for all measurable open subsets $U$

*$\lim_{n \to \infty}\mu_n(A) = \mu(A)$ for all measurable $A$ with $\mu(\partial A) = 0$
 A: Let's momentarily ignore 3. and 4.. 
Hopefully you agree that any notion of convergence of distributions $\mu_n$ to a distribution $\mu$ would imply $\mu_n(A) \rightarrow \mu(A)$ for an appropriate class of $A$. In this case the equivalence of 1. and 5. should be natural. The requirement that $\mu(\partial A) = 0$ allows us to say $\delta_{1/n} \Rightarrow \delta_0$, where $\delta_x(dx)$ is the measure $\delta_x(A) = 1$ if $x \in A$ and $0$ otherwise. 
Regarding 2., you should poke this definition for yourself a little bit. Compare with the definition of weak-$*$ convergence on the Banach space $X = C(S)$ for $S = [0, 1]$ and $S = [0, \infty)$ to get a feeling of this "weak" formulation of convergence. Keep in mind that the space of probability measures $\mathcal{P}(S)$ on a Polish space $S$ is not a vector space as it is not closed under addition. An equivalent statement of 2. is $\int f d\mu_n \rightarrow \int f d\mu$ for $f$ only bounded and continuous ($f\in C_b(S)$), and one can even relax this to $f \in C_b(S)$ which additionally vanish at infinity (aka $f \in C_0(S)$) under the assumption that $\mu$ is itself a probability measure (consider $\mu_n = \delta_n$). These puzzles are, as others have noted, up to you to think about.
If Portmanteau were just 1., 2. and 5., I suspect you would not have posted this question. What is confusing is that you can get away with the apparently "weaker" inequalities 3. and 4., and I interpret your question as "why am I not actually giving anything up?".
First, note that 3. and 4. are directly equivalent, simply by taking complements. Then, note that 3. and 4. together imply for any set $A$,
$$ \mu(A^\circ) \leq \liminf_n \mu_n(A) \leq \limsup_n \mu_n(A) \leq \mu(\bar{A}), $$
where $A^\circ$ and $\bar{A}$ are the interior and closure of $A$, respectively. In particular, if $\mu(\bar{A}) = \mu(A^\circ)$, then $\mu_n(A) \rightarrow \mu(A)$. If you recall that $\bar{A} = A^\circ 
\cup \partial A$, you recover 5. 
That is the "formal" explanation, but I would say you should look deeper at semicontinuity definitions. Consider, for instance, that a lower semicontinuous function has closed level sets and acheives its minimum on compact sets. These definitions are strong enough to guarantee some notion of convergence, but flexible enough that they extend to general contexts. Take a look at  $\Gamma$-convergence or the Large deviations principle. 
My point is these inequalities are useful to keep in mind in their own right, and the things to think about are how you would approximate the indicator function of an open set or a closed set. This is how the Portmanteau theorem is usually proved, and the best reference for weak convergence in general is Billingsley's book Convergence of Probability Measures.
A: This is a subjective question; I assume all answers are subjective.
My take is, weak convergence of  measures is such a natural concept that is has been rediscovered many times, and formalized slightly differently by each rediscover.  The Portmanteau theorem is says, in effect, all these definitions are equivalent.  
