Is this a valid identity for the expected value of a nonnegative random variable? Say $Z$ is a nonnegative random variable. First we can describe the integer component of $Z$ as $\sum_{n\in\mathbb{N}} I_{n\leq Z}$. Since $Z$ is always greater or equal to its integer component we have
$$
\begin{split}
\mathbb{E}[Z]
& \geq \mathbb{E}\left[\sum_{n\in\mathbb{N}} I_{n\leq Z}\right]\\
& = \sum_{n\in\mathbb{N}}\mathbb{E}\left[I_{n\leq Z}\right]\\
& = \sum_{n\in\mathbb{N}}\mathbb{P}\left(Z\geq n\right).
\end{split}
$$
On the other hand we can also deduce the following.
$$
\begin{split}
\mathbb{E}[Z]
& = \int_\Omega Z\,d\mathbb{P}\\
& = \int_\Omega \int_{x=0}^Z dx\,d\mathbb{P}\\
& = \int_\Omega \int_{\mathbb{R}^+} I_{[0, Z]}\,dx\,d\mathbb{P}\\
& = \int_{\mathbb{R}^+} \int_\Omega I_{[0, Z]}\,d\mathbb{P}\,dx\\
& = \int_{\mathbb{R^+}}\mathbb{P}\left(Z \geq x\right)\,dx
\end{split}
$$
These results I've seen in the literature, so I don't think I'm off so far. But now use the fact that $\mathbb{P}\left(Z \geq x \right)$ is a decreasing function of $x$ to write
$$
\begin{split}
\mathbb{E}[Z]
& = \int_{\mathbb{R^+}}\mathbb{P}\left(Z \geq x\right)\,dx\\
& = \sum_{n\in\mathbb{N}}\int_{x=n}^{n+1}\mathbb{P}\left(Z \geq x\right)\,dx\\
& \leq \sum_{n\in\mathbb{N}}\mathbb{P}\left(Z \geq n\right).
\end{split}
$$
Finally, putting the first and last result together we arrive at
$$
\mathbb{E}[Z] = \sum_{n\in\mathbb{N}}\mathbb{P}\left(Z \geq n\right).
$$
I'm a little perplexed that this result would hold for arbitrary nonnegative random variables. Is it actually correct?
 A: I see where the mistake lies. First you say the integer part of $Z$ is $\sum_{n\in\mathbb{N}} I_{\{n\leq Z \}}$ and this is only valid if $0\notin\mathbb{N}$. Alright? Now in your last lines you should have written the following:
\begin{align}
E[Z]&=\int_{\mathbb{R}^+}P(Z\geq x)\,dx\\
&=\color{red}{\sum_{n\in\mathbb{N}}\int_{n-1}^nP(Z\geq x)\,dx}\\
&\leq \sum_{n\in\mathbb{N}}P(Z\geq n-1)
\end{align}
But then you cannot conclude what you have concluded first. 
Actually I knew already that it is not true, but I was looking for the mistake. Take for example $Z\sim \exp(1)$ then $E[Z]=1$ but then the expresssion: $$\sum_{n\in\mathbb{N}} P(Z\geq n) =\sum_{n\in\mathbb{N}} e^{-n}\neq 1$$ in either cases $0\in\mathbb{N}$ or $0\notin\mathbb{N}$.
Edit. You have commented that $0\in\mathbb{N}$. Then, please, be consistent with everything, because the integer part of $Z$ would be $\sum_{n\in\mathbb{N}}I_{\{n+1\leq Z\}}$
A: You're right to be perplexed. This would seem to imply that $E(\lfloor Z \rfloor) = E(Z) $ for all non-negative random variables. Say $Z$ is an exponential random variable. Then we have $$ \int_0^\infty P(Z\ge x) dx = \int_0^\infty e^{-\lambda x}dx = \frac{1}{\lambda}=E(Z)$$ as promised. But we have $$ \sum_{n=0}^\infty P(Z\ge n) = \sum_{n=0}^\infty e^{-\lambda n } = \frac{1}{1-e^{-\lambda}}$$ which is generally larger than $1/\lambda.$ 
So in fact your second inequality $$ E(Z) \le \sum_{n=0}^\infty P(Z\ge n)$$ seems to be the correct one.
What was wrong with the first one? After all, it came from the seemingly bulletproof argument that $$\lfloor Z\rfloor=\sum_{n=0}^\infty 1(Z \ge n).$$ Except, wait... it's not. We have $$ 1 = \lfloor 1.1 \rfloor \ne 1+ 1 = 2.$$ This is overcounting by one. So in fact your first inequality should be $$E(Z) \ge -1+\sum_{n=0}^\infty P(Z\ge n).$$
This, perhaps more than anything, underscores the importance of never writing $$ \sum_{n\in \mathbb N} f(n)$$ and always writing $ \sum_{n=0}^\infty f(n)$ or $ \sum_{n=1}^\infty f(n),$ whichever is appropriate.
