Lévy's Inversion Formula

Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$.

Let $ F_X$ be the distribution function of $X$.

Let $\phi : \R \to \C$ be the characteristic function of $X$.

Then for all $a < b$ such that:
 * $\map \Pr {X \in \set {a,b} } = 0$

we have:

Proof
In fact, the first equality holds for all $a < b$, as:

In the following, we shall show the second equality.

Let $\mu$ be the probability distribution of $X$.

Let $m$ be the Lebesgue measure on $\R$.

For $T > 0$ let $m_T$ be the restriction of $m$ to $\closedint {-T} T$, i.e.:
 * $\forall A \in \map \BB \R : \map {m_T} A := \map m { A \cap \closedint {-T} T}$

Let $\mu \times m_T$ be the product measure.

Then:

where:
 * $\ds \map f {x,t} := \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it}$

The $f$ is essentially bounded with respect to $\mu \times m_T$, since:
 * $\forall \struct {x, t} \in \R \times \R_{\ne 0} : \cmod {\dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it} } \le \dfrac 3 2$

by Bounds for Complex Exponential.

In particular, $f$ is $\mu \times m_T$-integrable, as $\mu \times m_T$ is finite.

Thus by Fubini's Theorem:

where:

Observe:

where:
 * $\sgn : \R \to \set {-1, 0, 1}$ is the signum function
 * $\Si : \R \to \R$ is the sine integral function

Similarly:
 * $\ds \int _{-T}^T \dfrac {e^{it \paren {x - b} } }{it} \rd t = 2 \; \map \sgn {x - b} \; \map \Si {T \size {x - b} }$

Thus we have:
 * $\ds \map {F_T} x = 2 \; \map \sgn {x - a} \; \map \Si {T \size {x - a} } - 2 \; \map \sgn {x - b} \; \map \Si {T \size {x - b} }$

By Limit at Infinity of Sine Integral Function, for all $x \in \R \setminus \set {a,b}$:

As $\map \mu {\set {a, b} } = 0$, this means in particular:
 * $\ds \lim _{T \mathop \to +\infty} F_T = \chi _{\hointl a b}\quad$ $\mu$-almost surely

On the other hand, Sine Integral Function is Bounded:
 * $\ds \sup_{T > 0, \; x \in \R} \size {\map {F_T} x} \le 4 \norm \Si_\infty < +\infty$

Note that the constant function $4 \norm \Si_\infty$ is $\mu$-integrable.

Therefore: