Theorem: Lévy's Inversion Formula

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:
{{begin-eqn}}
{{eqn | l = \map {F_X} b - \map {F_X} a
      | r = \map \Pr {a < X \le b}
}}
{{eqn | r = \lim _{T \to \infty} \frac 1 {2 \pi} \int^T _{-T} \dfrac {e^{-ita} - e^{-itb} }{it} \map \phi t \rd t
}}
{{end-eqn}}

Proof:
In fact, the first equality holds for all $a < b$, as:
{{begin-eqn}}
{{eqn | l = \map {F_X} b - \map {F_X} a
      | r = \map \Pr {X \le b} - \map \Pr {X \le a}
      | c = {{Defof|Cumulative Distribution Function|$F_X$}}
}}
{{eqn | r = \map \Pr {\set {X \le b} \setminus \set {X \le a} }
      | c = Measure of Set Difference with Subset, as $\set {X \le a} \subseteq \set {X \le b}$
}}
{{eqn | r = \map \Pr {a < X \le b}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \int^T _{-T} \dfrac {e^{-ita} - e^{-itb} }{it} \map \phi t \rd t
      | r = \int^T _{-T} \int \dfrac {e^{-ita} - e^{-itb} }{it} e^{i t x} \rd \map \mu x  \rd t
}}
{{eqn | r = \iint \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it} \rd \map \mu x \rd \map {m_T} t }}
{{eqn | r = \iint \map f {x,t} \rd \map \mu x \rd \map {m_T} t
      | c = $(1)$
}}
{{end-eqn}}
where:
:$\ds \map f {x, t} := \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } } {i t}$
The $f$ is essentially bounded with respect to $\mu \times m_T$ , since:
:$\forall \struct {x, t} \in \R \times \R_{\ne 0} : \cmod {\dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } } {i t} } \le \dfrac 3 2$
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:
{{begin-eqn}}
{{eqn | l = (1)
      | r = \int f \; \rd \paren {\mu \times m_T}
}}
{{eqn | r = \iint \map f {x,t} \rd \map {m_T} t \rd \map \mu x 
}}
{{eqn | r = \int F_T \rd \mu 
}}
{{end-eqn}}
where:
{{begin-eqn}}
{{eqn | l = \map {F_T} x
      | o = :=
      | r = \int \map f {x,t} \rd \map {m_T} t
}}
{{eqn | r = \int _{-T}^T \dfrac {e^{it \paren {x - a} } - e^{it \paren {x - b} } }{it} \rd t
}}
{{eqn | r = \int _{-T}^T \dfrac {e^{it \paren {x - a} } }{it} \rd t - \int _{-T}^T \dfrac { e^{it \paren {x - b} } }{it} \rd t
}}
{{end-eqn}}
Observe:
{{begin-eqn}}
{{eqn | l = \int _{-T}^T \dfrac {e^{it \paren {x - a} } }{it} \rd t
      | r = \int _{-T}^T \dfrac {\map \cos {t \paren {x-a} } + i \map \sin {t \paren {x-a} } }{it} \rd t
      | c = Euler's Formula
}}
{{eqn | r = \int _{-T}^T \dfrac {\map \sin {t \paren {x-a} } }{t} \rd t
      | c = Definite Integral of Odd Function
}}
{{eqn | r = 2 \int _0^T \dfrac {\map \sin {t \paren {x - a} } } t \rd t
      | c = Definite Integral of Even Function
}}
{{eqn | r = 2 \; \map \sgn {x - a} \int _0^T \dfrac {\map \sin {t \size {x - a} } } t \rd t
      | c = Sine Function is Odd
}}
{{eqn | r = 2 \; \map \sgn {x - a} \int _0^{T \size {x-a} } \dfrac {\sin s } s \rd s
      | c = Integration by Substitution with $s = t \size {x-a}$
}}
{{eqn | r = 2 \; \map \sgn {x - a} \; \map \Si {T \size {x - a} }
}}
{{end-eqn}}
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 $x \in \R \setminus \set {a,b}$:
{{begin-eqn}}
{{eqn | l = \lim _{T \mathop \to +\infty} \map {F_T} x
      | r = \pi \paren {\map \sgn {x - a} - \map \sgn {x - b} }
}}
{{eqn | r = 2 \pi \; \map {\chi _{\hointl a b} } x
}}
{{end-eqn}}
As $\map \mu {\set {a, b} } = 0$ {{hypothesis}}, 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:
{{begin-eqn}}
{{eqn | l = \lim _{T \mathop \to +\infty} \int F_T \rd \mu
      | r = \int \lim _{T \mathop \to +\infty} F_T \rd \mu
      | c = Lebesgue's Dominated Convergence Theorem
}}
{{eqn | r = 2 \pi \int \chi _{\hointl a b} \rd\mu
}}
{{eqn | r = 2 \pi \; \map \mu {\hointl a b}
}}
{{eqn | r = 2 \pi \; \map \Pr {X \in \hointl a b}
      | c = {{Defof|Probability Distribution/Real-Valued Random Variable|Probability Distribution}}
}}
{{eqn | r = 2 \pi \; \map \Pr {a < X \le b}
}}
{{end-eqn}}
{{finish|Some details}}
{{Namedfor|Paul Pierre Lévy|cat = Lévy}}