Which statistic is this 'overlap between sets/sections' - if any? I’m sure someone in history already invented the following formula before.  Does anyone know if so and what it’s called?
Let’s say there is a set of number pairs, such that the second number in each pair is always greater than or equal to the first number.  We’ll call the first number in each pair the “lower” and the second number in each pair the “upper”.  For example:
[3-9], [5-12], [1-4], [6-17], [4-5]
So the first element in set is [3-9], which means its lower is 3 and its upper is 9.  In other words, it’s covering the range 3 to 9.
Now let’s calculate a new statistic that I just thought up.  Let’s call it the Overlap Score.  Here’s how we would calculate it.  (Pardon the psuedo-code; I’m not as familiar with mathematical notation.)
   For i = 1 to n elements in the set
      For j = (i + 1) to n elements in the set
         We compare the (i)th pair with the (j)th pair
         If i(lower) > j(upper) or i(upper) < j(lower)
            Then i and j don’t overlap

So we would sum (sigma or Σ) the number of overlaps and then divide by the total number comparisons we made (which is N choose 2).
And for those of you familiar with Python, here’s the Python version:
pairs = [(3,9), (5,12), (1,4), (6,17), (4,5)]

overlaps = 0
comparisons = 0
for pair in pairs:
    for pair2 in pairs:
        if pairs.index(pair2) <= pairs.index(pair):
            continue
        comparisons += 1
        overlap = False
        if not (pair[0] > pair2[1] or pair[1] < pair2[0]):
            overlaps += 1
            overlap = True
        print(pair, "overlaps", pair2, "=", overlap)
print()
print ("Score:", overlaps, "/", comparisons, "=", (overlaps / comparisons))

So let’s calculate the example above:
(3, 9) overlaps (5, 12) = True
(3, 9) overlaps (1, 4) = True
(3, 9) overlaps (6, 17) = True
(3, 9) overlaps (4, 5) = True
(5, 12) overlaps (1, 4) = False
(5, 12) overlaps (6, 17) = True
(5, 12) overlaps (4, 5) = True
(1, 4) overlaps (6, 17) = False
(1, 4) overlaps (4, 5) = True
(6, 17) overlaps (4, 5) = False

Score: 7 / 10 = 0.7

So that makes a total of 7 overlaps out of 10 combinations.  So the final score would be: 0.7
So a score of 1.0 would be where every pair overlaps with every other pair.  And a score of 0.0 would be where no pair overlaps with any other pair.
And the point of this would be to look at a set of ranges and calculate a score that tells us how “overlapping” the set is.  In particular, I would use it for a sets of values in a box plot that have highs and lows based on margins-of-error.
 A: I am not sure whether there is a specific description for this problem, but one analogous way to view your problem is as an adjacency matrix
where the entries are 0 or 1 depending on the overlap (you can view it as a graph with edges/neighbors depending on the overlap). 
$$\begin{array}{cccccccc}
& [3-9] & [5-12] & [1-4] & [6-17] & [4-5] &\\
\begin{array}{}
  [3-9] \\ [5-12] \\ [1-4] \\ [6-17] \\ [4-5]
\end{array}
\begin{array}{}
  \left( \vphantom{\begin{array}{}  [3-9] \\ [5-12] \\ [1-4] \\ [6-17] \\ [4-5] \end{array}} \right.
\end{array}
& \begin{array}{}  - \\ 1 \\ 1 \\ 1 \\ 1 \end{array}
& \begin{array}{}  1 \\ - \\ 0 \\ 1 \\ 0 \end{array}
& \begin{array}{}  1 \\ 0 \\ - \\ 0 \\ 0 \end{array}
& \begin{array}{}  1 \\ 1 \\ 0 \\ - \\ 0 \end{array}
& \begin{array}{}  1 \\ 0 \\ 0 \\ 0 \\ - \end{array}& 
\begin{array}{}
  \left. \vphantom{\begin{array}{}  [3-9] \\ [5-12] \\ [1-4] \\ [6-17] \\ [4-5] \end{array}} \right)
\end{array} 
\end{array}$$
And then you are interested in the average degree of the nodes (or you can look into more detail, for instance the distribution of the degree of the nodes).
Here the nodes [3-9], [5-12], [1-4], [6-17], [4-5] have respectively degree 4,2,1,2,1 and the mean degree is 2. The maximum possible degree is 4. And the ratio of the average degree with the maximum possible degree is 0.5 (you had counted one too many, "[1-4]... overlaps [4-5]")
