probability - Expectation of non-negative random variable

Let $X$ be a non-negative random variable. In a proof for $E[X]=\int_0^\infty P(X>t)dt$ from the answer of this question, we use Fubini for the middle quality. Why do we need $X$ to be non-negative? We basically have a double integral over a function $f(X,t)$ which is $1$ if $X>t$ and else 0. So this function is non-negative for any $X$ not just for non-negative random variables $X$, thus we could use Fubini regardless. Where is the flaw in my reasoning?

Answer

We require $X$ to be non-negative, because otherwise
\begin{equation}

X \neq \int_{[0,X)} 1 \ dt, \ \ \mbox{ so in general} \ \ E[X] \neq E \left ( \int_{[0,X)} 1 \ dt \right ).
\end{equation}

For example, if for some $\omega_0 \in \Omega$ we have that $X(\omega_0) = -5 < 0$, then $[0,X(\omega_0)) = \emptyset$ and $\int_{[0,X(\omega_0))} 1 \ dt = 0 \neq X(\omega_0)$, so we are in trouble if $\mathbb{P} (\omega_0) >0$. Thus, $X$ is required to be non-negative in order for the first equality to hold, not because of the Fubini theorem.

Not that for a non-positive $X$ you get
\begin{equation}
E[X] = -E[-X] =- \int_0^{\infty} P (- X > t)\ dt= \int_{-\infty}^0 P (X \end{equation}