Let X be a non-negative random variable. In a proof for E[X]=∫∞0P(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
X≠∫[0,X)1 dt, so in general E[X]≠E(∫[0,X)1 dt).
For example, if for some ω0∈Ω we have that X(ω0)=−5<0, then [0,X(ω0))=∅ and ∫[0,X(ω0))1 dt=0≠X(ω0), so we are in trouble if P(ω0)>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}
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