I am having trouble getting past an equality given in a proof in Cinlar "Probability and Stochastics"
In theorem 2.3.19, he makes the following immediate assertion, where $X$ is a positive, $\mathbb{R}$-valued random variable in a uniformly integrable collection $\mathcal{K}$ and $1_A$ is the characteristic function on set $A$.
$\mathbb{E} ~ X \cdot 1_{\{X > b\}} = \int_0^\infty dy ~ \mathbb{P} \{X \cdot 1_{\{X > b\}} > y\}$
So far, expectation has been defined as
$\mathbb{E} X = \int_\Omega \mathbb{P} (d\omega) X(\omega)$
And we have a theorem
$\mathbb{E} ~ (f \circ X) = \int_\mathbb{R} P \circ X^{-1}(dx) f(x)$
Yet I fail to use either of these to produce the Lebesgue measure (to get a Riemann integral) on the RHS above, and (without a specific distribution defined as such) I don't really understand why we get a Lebesgue measure necessarily.
So letting $f(x) =x \cdot 1_{(b, \infty]}(x)$ then $X \cdot 1_{\{X > b\}} = f \circ X$ so
$\mathbb{E} ~ X \cdot 1_{\{X > b\}} = \mathbb{E} (f \circ X)$
$= \int_\mathbb{R} \mathbb{P} \circ X^{-1}(dx) f(x)$
$= \int_\mathbb{R} \mathbb{P} \circ X^{-1}(dx) x \cdot 1_{(b, \infty]}(x)$
Where can I go from here to get to the RHS above?
Best Answer
I believe you can obtain the desired equality using the observation that $$Y=\int^Y_0dy=\int_0^{\infty}\mathbb{1}_{\{Y\geq y\}}dy$$ Taking expectations on both sides we obtain, $$\mathbb{E}Y=\int_0^{\infty}\mathbb{P}(Y\geq y)dy$$ Source: Lieb and Loss "Analysis."
Let $Y=X1_{X\geq b}$. Then by applying the above result to $Y$, we obtain $$X1_{X\geq b} = \int_0^{\infty}\mathbb{P}(X1_{\{X>b\}}\geq y)dy$$