Prove $\mathbb E(f(Y)) = \mathbb E\Bigl (\int_0^\infty 1_{\lambda \le Y} df(\lambda)\Bigr)$

Question

This is part of a proof in a book i'm reading which states the following

Show that for $Y = \sup_{s≤t} X_s$ and $Z = X_t = X_{t}^+$, then we know for an increasing, right-continuous function $f$ such that $f(0) = 0$, then:

$$ \mathbb E(f(Y)) = \mathbb E\Bigl (\int_0^\infty 1_{\lambda \le Y} df(\lambda)\Bigr)$$

Can anyone explain why the above holds? The notes make it seem trivial as there is no explanation but does not seem trivial to me (i.e. what is $\lambda$) and i feel a separate proof is needed, which i'm not sure how to prove.

Answer 1

$f$ is increasing and right continuous so by Lebesgue–Stieltjes integration $$f(b)-f(a) = \int_{a}^{b} df(\lambda) = \int_{-\infty}^{\infty} 1_{a\leq f(x)\leq b}df(\lambda) $$ In particular $Y$ is positive, then points where $Y\leq 0$ has zero measure and you'll get $$f(Y)= f(Y)-f(0) = \int_{0}^{\infty} 1_{0\leq \lambda\leq Y} df(\lambda)$$