Prove that $\mathbb{E}|X| < \infty \Longleftrightarrow \int_0^1 F_X^{-1}(t)~ \lambda(dt) < \infty$

Question

Let $X$ be a random variable with probability distribution function $F_X$. Its inverse function $F_X^{-1}$ is defined as: $$F_X^{-1}(t) = \text{inf} \left( x : F_X(x) \leqslant t \right) \quad;\quad t \in [0,1].$$

Prove that $$\mathbb{E}|X| < \infty \quad \Longleftrightarrow \quad \int_0^1 F_X^{-1}(t)~ \lambda(dt) < \infty$$ where $\lambda$ denotes the Lebesgue measure.

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An apparently correct solution involves the substitution $t = F_X(z)$. But since we don't know much about $F_X$, I don't think that's a correct approach. Any help will be appreciated.

Answer 1

You have to switch $\leq$ into $\geq$ in the definition of $F^{-1}$. Further: Show it via the following approach:

1. First assume that $X$ is nonnegative
2. Show it for indicator functions of intervals
3. Show that the set of functions $h$ for which the equality $\int_0^\infty h \mathrm{d}F=\int_0^1 h(F_X^{-1}(t))1_{F_X^{-1}<\infty}\mathrm{d}t$ is true, satisfies the requirements of the monotone class theorem: https://en.wikipedia.org/wiki/Monotone_class_theorem

Edit: You can start with e.g.: $$\int 1_{(a,b]}(y)\mathrm{d}F_X(y) = F(b)-F(a)$$ which equals: $$\int_0^1 1{(a,b]}(F_X^{-1}(t))\mathrm{d}t$$

After using the monotone class theorem you will have shown it for the function $x \mapsto x$ Which will give something like (provided X is nonnegative): $$\mathbb{E}(X) = \int x\mathrm{d}F_X(y) = \int_0^1 F_X^{-1}(t)\mathrm{d}t$$

Now the final step is achieved via decomposing $X$ into $X=X^+-X^-$