Is sum of tail probability always less than integral of tail probability? I'm working through some Probability and Measure Theory, and frequently we have been using the fact that for $X_i$  iid
$\sum\limits_{k = 1}^{\infty} P(|X_1| > k) \leq \int\limits_0^{\infty}P(|X_1| >t)dt$
Intuitively this makes sense but other than seeing a graph, I haven't been able to convince myself why this is always true, nor have I been able to come up with a counter example. Any insights on how to prove/disprove for a generic iid r.v?
Thanks!
 A: If you have two events, $|X|>k_1$ and $|X|>k_2\geq k_1$, The second event includes the first, so $P(|X| > k_1) \leq P(|X| > k_2)$
In particular, $P(|X|>k) \leq \int_k^{k+1} P(|X| > k) dt$ for any natural number $k$.
A: You have $\int_0^1 P(X_1\gt t) dt \ge \int_0^1 P(X_1\gt 1)dt=P(X_1 \gt 1)$ and similarly  $\int_n^{n+1}1 P(X_1\gt t) dt \ge \int_0^1 P(X_1\gt n)dt=P(X_1 \gt n)$, so each interval in the integral dominates the element in the sum at the end.
A: Actually the inequality goes the other way round:
$$\sum_{n=0}^\infty \Bbb P(|X|>n) ≥ \int_0^\infty \Bbb P(|X|>t)dt$$
Here’s a quick sketch of a proof:
Let p and q be natural numbers s.t. p ≥ q. 
Then, for any random variable X, it is easier for |X| to be larger than q than to be larger than p, since q is smaller. 
So if |X| is larger than p, then a fortiori |X| is larger than q, hence the event {|X|>p} is included in the event {|X|>p}.
(i.e. the set of outcomes that make |X| bigger than p also make |X| bigger than q)
Thus we find the inequality 
(1) $ \Bbb P(|X|>p) ≤ \Bbb P(|X|>q)$
It now follows that 
(2) $ \Bbb P(|X|>n) ≥ \int_n^{n+1} \Bbb P(|X|>t)dt $
Hence by taking the sum over all natural numbers on both sides of the last inequality, we find
(3) $\sum_{n=0}^\infty \Bbb P(|X|>n) ≥ \sum_{n=0}^\infty \int_n^{n+1} \Bbb P(|X|>n) = \int_0^\infty \Bbb P(|X|>t)dt$
and the claim follows.
In addition, it is sometimes useful in this context to know that the following result holds:
(4) “The sum $\sum \Bbb P(|X|>p)$ and the integral $\int \Bbb P(|X|>p)$ are of the same kind”
in the sense that
$\sum \Bbb P(|X|>p)$ is finite iff $\int \Bbb P(|X|>p)$ is finite, and 
$\sum \Bbb P(|X|>p)$ is infinite iff $\int \Bbb P(|X|>p)$ is infinite
Note the the lower bound of summation/integration plays no role in the finiteness of these sums/integrals, as long as we sum/integrate up to infinity.
To see that (4) holds, consider equation (2) with the following additional lower bound:
(5) $\Bbb P(|X|>n) ≥ \int_n^{n+1} \Bbb P(|X|>t)dt ≥ \Bbb P(|X|>n+1)$
which follows directly from our preliminary observations.
By taking the infinite sum over all natural n in (5) one finds
(6) $\sum_{n=0}^\infty \Bbb P(|X|>n) ≥ \int_0^\infty \Bbb P(|X|>t)dt ≥ \sum_{n=0}^\infty \Bbb P(|X|>n+1)  = \sum_{n=1}^\infty \Bbb P(|X|>n)$
The right and left most sum only differ by the single term $\Bbb P(|X|>0)$ which is an element of $[0,1]$ and thus in particular finite. Therefore either both sums are finite or infinite. 
Hence any of the two sums is finite if the integral is finite (using the second inequality), and the sums are infinite if the integral is infinite (using the first inequality).
This shows the claim in (4).
