O(1) space solution.

import random
import pprint

def interval_random(intervals):
   # find how many numbers
   total = 0

   for interval in intervals:
       total += interval[1] - interval[0] + 1
   # get random sequence number
   seq = random.randint(0, total-1)

   # go through intervals again to cherry pick the original integer
   # corresponding to the global sequence number
   for interval in intervals:
       if seq < interval[1] - interval[0] + 1:
           return seq + interval[0]
       # here you only need to offset the current seq by length of the
       # last interval
       seq -= interval[1] - interval[0] + 1
   return seq + intervals[-1][0]

if __name__ == "__main__":
   intervals = [(7,9),(3, 5), (10, 18), (20, 26), (30, 32)]
   print(interval_random(intervals))

   # uniform distribution test
   counts = {}

   for i in xrange(0, 2500000):
       result = interval_random(intervals)
       if result not in counts:
           counts[result] = 1
       else:
           counts[result] += 1
   print(len(counts))
   pp = pprint.PrettyPrinter(depth=6)
   pp.pprint(counts)

"""
Output:
21
25
{3: 100269,
4: 100183,
5: 99901,
7: 99685,
8: 100031,
9: 100330,
10: 99832,
11: 100394,
12: 99429,
13: 99923,
14: 100082,
15: 100060,
16: 100049,
17: 100479,
18: 100327,
20: 99947,
21: 99986,
22: 99299,
23: 100115,
24: 100104,
25: 99638,
26: 99741,
30: 100056,
31: 100045,
32: 100095}
"""