Proving $\sum_\limits{n=1}^{\infty}\mathbb{P}(X\geqslant n)\leqslant E(X)\leqslant 1+\sum_\limits{n=1}^{\infty}\mathbb{P}(X\geqslant n)$ 
Let $(\Omega,\mathscr{A},\mathbb{P})$ a probability space and $X:\Omega\to\mathbb{R}$ a random variable. Prove:
$$\sum_\limits{n=1}^{\infty}\mathbb{P}(X\geqslant n)\leqslant E(X)\leqslant 1+\sum_\limits{n=1}^{\infty}\mathbb{P}(X\geqslant n)$$

I have an intuition for the sums when $X$ is discrete but I do not know how to prove the statement. I have no idea. I need to get a grip on how to tackle problems with probability measure.
Question:
Can someone help me prove the inequalities?
Thanks in advance!p
 A: This is true only if $X$ is non-negative. 
Let $Y = \lfloor X \rfloor$ (floor function, integer part). Then $Y \le  X < Y+1$ and
$$ E[Y] \le E[X] < 1+E[Y] \tag1$$
Also, if $Y$ is a discrete non-negative random variable, then $$ E[Y] = \sum_{n=1}^\infty  \mathbb{P}(Y\geqslant n)\tag2$$
Further, for any integer $n\ge 1$, $X \ge n \iff  Y \ge n$ 
Then 
$$
\sum_\limits{n=1}^{\infty}\mathbb{P}(X\geqslant n)\leqslant E(X) <1+\sum_\limits{n=1}^{\infty}\mathbb{P}(X\geqslant n) \tag3$$
which is a little stronger than the original inequality.
A: I am assuming $X$ is non-negative, i.e., $X\colon \Omega\to [0,\infty)$. Otherwise, the result you aim to prove is false.
Hint: You can artificially "discretize" the integral defining expectation as follows:
$$
\mathbb{E}[X] = \int_0^\infty \mathbb{P}\{ X\geq x\} dx
= \sum_{n=0}^\infty \int_n^{n+1} \mathbb{P}\{ X\geq x\} dx \tag{1}
$$
and then use the easy bounds
$$
\mathbb{P}\{ X\geq n+1\} \leq \int_n^{n+1} \mathbb{P}\{ X\geq x\} dx\leq \mathbb{P}\{ X\geq n\} \tag{2}
$$
A: Let $Y=\lfloor X\rfloor$. Obviously, $EY\le EX$, because $0\le Y\le X$ always. Furthermore,
$$
E[Y]=\sum_{n=0}^\infty nP(Y=n)=\sum_{n=0}^\infty nP(n\le X< n+1)
$$
Let $p_n=P(n\le X<n+1)$. We will rearrange the summation as follows:
\begin{array}{cccccccccccccl}
\sum_n np_n= & p_1&+&p_2&+&p_3&+&\dots &+&p_n&+\dots &= &P(X\ge 1)& +
 \\& &&p_2&+&p_3&+&\dots &+&p_n&+\dots & &P(X\ge 2)& +
 \\& &&&&p_3&+&\dots &+&p_n&+\dots &&P(X\ge 3)& +
 \\& &&&& &&\ddots&&&\vdots &&\vdots
\\\\&&&&&&&&&&&=&\sum_{n=1}^\infty P(X\ge n)
\end{array}
Therefore, we see that
$$
E[X]\ge E[Y]=\sum_{n=1}^\infty np_n=\sum_{n=1}^\infty P(X\ge n).
$$
For the other inequality, let $Z=\lceil Z\rceil$, and use $EX\le EZ$.
