At least h with at least h

Python, 52

f=lambda s,n=0:n<sum(n<x for x in s)and f(s,n+1)or n

A recursive solution. Run this in Stackless Python if you're worried about overflows.

Starting from n=0, checks whether at least n+1 of the numbers are at least n+1. If so, increments n and starts again. If not, outputs n.

The conditional is done using Python's short-circuiting for Booleans. The expression sum(n<x for x in s) counts the number of values in s that are greater than n by adding the indicator Booleans, which are treated as 0 or 1.

For comparison, the iterative equivalent is 2 chars longer. It requires Python 2.

s=input()
n=0
while n<sum(n<x for x in s):n+=1
print n

Unfortunately, the input need to be saved for a variable before being iterated over or else Python will try to read the input repeatedly.

Pyth, 13 10 bytes

tf<l-QUTT1

Input in a form such as [22,33,1,2,4] on STDIN.

Try it here.

How it works:

-QUT is all of the numbers in the input (Q) at least as large as the number being checked, T.

<l-QUTT is true if the the length of that list is less than T.

f<l-QUTT1 finds the first integer which returns true for the inner check, starting at 1 and going up.

tf<l-QUTT1 decrements that by one, giving the largest value for which the condition is false, which is the h-index.

Starting at 1 ensures that 0 is returned when the test is always true, such as in the first test case.

Python 2, 49

Input should be typed in the same format as the examples.

i=0
for z in sorted(input())[::-1]:i+=z>i
print i