Q:

Are (functions of) sets naturally continuous in some way which real numbers aren't? 

I am reading some introductory notes on probability theory, and I came across the following claim here (page 7/8), :
If some sequence of sets $A_n$ has the limit $A$, then
$\lim_{n \rightarrow \infty} \mu(A_n) = \mu(A)$ for any measure $\mu$.
This really surprised me because it looks just like the definition for continuous functions (of real numbers), and it's not immediately apparent why all measures should be "continuous".  The fact that it's possible to have a discontinuous cumulative distribution function makes this even more unintuitive. 
It seems that one property of sets which makes this work is that the limit of a sequence of real numbers can lie outside the "range" of that sequence (for example, $a_n = 1/n$ has limit $L=0$), whereas if i take $A_n$ to be the interval $(a_n, \infty)$ or  $[a_n, \infty)$, I get $(L, \infty)$ as the limit, which doesn't contain the limit ($0$) of $a_n$. Otherwise, one could set $\mu(\{0\}) = \infty$ and cause a discontinuity. 
My question is 1).  is there some obvious intuition I'm missing with respect to how unintuitive this seems?  2).  Are sets/real numbers a special case, or do other objects also naturally "cause" functions to be continuous? 

A:

The measure of a set captures the abstract notion of "how big a set is".  Measure theory deals with the construction of measures and the proofs of their properties.  So the precise answer to your question can only be given once that background is properly established. 
In the absence of that, I'll try to convey some of the intuition behind these concepts.  Measures are closely related to integrals; If you have a definition of the integral, the measure of the set is just the integral of the indicator function of the set:
$$ \mu( (-1,3)) = \int_{-1}^3 1 \,dx = 4.$$
Conversely, you can define the integral if you start from a measure. 
From that point of view, it shouldn't be surprising that $\mu( (0,1) ) = \mu([0,1))$. Indeed, when you write the familiar (Riemann) integral, you don't differentiate between them!  This is because a single point is too "thin" with respect to the line and the area under a point - a line segment - is too "thin" with respect to area under a curve. 
Setting $\mu(\{0\}) =\infty$ can be done, but it needs a lot of extra care.  Doing this leads to the definition of the Dirac delta function, which isn't really a function, but a measure!  Besides this, giving any other "finite value" to $\mu(\{0\})$ doesn't change the integral. 
You can think of a function $f(x):[0,1]\to \mathbb{R}$ and a function $g$ with $f(x)=g(x)$ for $0\leq x <1$. Then
$$\int_0^1 f(x)dx= \int_0^1 g(x) dx,$$
as we learn in calculus, even though $f(1)\neq g(1)$. For that reason, when dealing with the "usual" measure in $\mathbb{R}$ (called the Lebesgue measure) we have $\mu(\{p\}) =0$ for every single point $p\in\mathbb{R}$. You can think of that as
$$\int_p^p f(x) dx = 0,$$
for every function $f$. That way $\mu ( [0,1]) = \mu ([0,1)) + \mu(\{1\}) = \mu ([0,1))$.
Finally, to answer your question, the property that you found unintuitive is called the continuity of a measure, and it roughly says that if you have (any) true measure and you are measuring a set, if you measure things that approximate the set, then you get the measure of the set. 
Of course "approximating the set" needs to be done in a "nice" way, that's why the sequence of sets that you take need to be "increasing" or "decreasing".  See the wiki article for details.