Proving the uniqueness of characteristic functions without using Fourier transform theory. I am wondering if there is a way to prove that if two random variables have the same characteristic function, they have the same probability distribution, without using Fourier Transform theory?
Here is what I got so far:
What we need to prove is this:
Let $X,Y$ be two real random variables. If
$$E\left[e^{iuX}\right]=E\left[e^{iuY}\right],$$
for every $u \in \mathbb{R}$ then we need to show the distribution for $X$ and $Y$ is the same. That the distribution is the same means that for every $B\in \mathcal{B}(\mathbb{R})$ we have
$$E[1_B(X)]=P(X\in B)=P(Y\in B)=E[1_B(Y)].$$
We see that it suffices to show that for any bounded complex function $f:(\mathbb{R},\mathcal{B}(\mathbb{R}))\rightarrow (\mathbb{C},\mathcal{B}(\mathbb{C}))$ we have
$$E\left[f(X)\right]=E\left[f(Y)\right].$$
I think we maybe can either use a convergence theorem, or a monotone class theorem to finish the proof, but I am not sure how to do it. Do you see how to finish the proof?
 A: The following result

Suppose that   $\mu$ and $\nu$ are complex measures (measures of finite variation) on $\mathscr{B}(\mathbb{R}^d)$. Then, $\mu=\nu$ iff
$\widehat{\mu}=\widehat{\nu}$.

can be proven through monotone class arguments. Let $\mathcal{M}$ be the collection of all functions of the form  $f_{\bf t}({\bf x})=\exp(i\boldsymbol{x}\cdot \boldsymbol{t})$, $\boldsymbol{t}\in\mathbb{R}^d$.  This is a
complex multiplicative family and contains $\mathbb{1}=f_{\boldsymbol{0}}$.   $\mathcal{M}$ is contained in the space of all bounded complex
valued Borel measurable functions $\mathcal{V}$. The later is a
complex vector space and a bounded class.
By the complex bounded class theorem, $\mathcal{V}$ contains all the
bounded complex valued $\sigma(\mathcal{M})$--measurable
functions. In particular, $\mathcal{V}$ contains  all functions of the form
$\mathbb{1}_B$, $B\in\sigma(\mathcal{M})$.
Since $\mu$ and $\nu$ coincide in $\mathcal{M}$, then by dominated
convergence, they also coincide in $\sigma(\mathcal{M})$.
Consider the maps $\gamma_{\bf t}({\bf x})={\bf t\cdot x}$, with
${\bf  t}\in\mathbb{R}^d$ and observe that they generate  $\mathscr{B}(\mathbb{R}^d)$. Since
\begin{align}
\gamma_{\bf t}({\bf x}) = {\bf t\cdot x}=-i\lim_nn(f_{{\bf t}/n}({\bf x})-f_{\bf 0}({\bf x})),
\end{align}
each  $\gamma_{\bf t}$ is  $\sigma(\mathcal{M})$-measurable. Therefore
$\sigma(\mathcal{M})=\mathscr{B}(\mathbb{R}^d)$ and $\mu=\nu$.

Here are more or less the tools that make this argument work:
A few definitions:

*

*A collection  $\mathcal{V}\subset\mathbb{R}^\Omega$, is a
monotone class if it is closed under taking pointwise limits of
monotone convergent sequences.

*A collection  $\mathcal{V}\subset\mathcal{B}_b(\Omega;\mathbb{R})$ is a bounded monotone class if it is  closed under taking pointwise limits of  uniformly bounded monotone sequences.

*A collection $\mathcal{V}\subset\mathcal{B}_b(\Omega;\mathbb{F})$, where $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$,   is
a bounded class if it is closed under taking pointwise limits of uniformly bounded convergent sequences.

*A collection  $\mathcal{M}\subset\mathbb{R}^\Omega$
is a real multiplicative class  if it is closed under finite multiplication.

*A collection  $\mathcal{M}\subset\mathbb{C}^\Omega$ of complex valued
functions is a
complex multiplicative class  if it is  closed under finite multiplication and under complex conjugation.

Theorem: (Complex bounded class theorem).  Suppose $\mathcal{V}\subset\mathcal{B}_b(\Omega;\mathbb{C})$ is a complex vector space,   a complex bounded class, and contains the constant function $\mathbb{1}$. If $\mathcal{M}\subset\mathcal{V}$ is complex multiplicative class, then $\mathcal{V}$ contains the collection of all bounded $\mathbb{C}$--valued $\sigma(\mathcal{M})$--measurable functions.
A: You may approximate $P(a<X<b)$ by a nondecreasing sequence of continuous functions $f_n$ such that $f(x)=0$ for $x\notin (a,b)$ and $f_n(t)\to I\{a<t<b\}$ for each $t$. Thus if $E[f_n(X)]=E[f_n(Y)]$ for all bounded continuous functions with compact support, $X$ and $Y$ have the same distributions by the monotone convergence theorem.
Now let us prove that $E[f(X)]=E[f(Y)]$ for all continuous $f$ with compact support. Assume WLOG that $f$ is supported on $[-b,b]$, and pick a large enough $a>|b|$ such that $P(|X|>a) + P(|Y|>a) \le \epsilon$. Then $f$ is also supported on the larger $[-a,a]$.
Now if $f:R\to C$ is continuous, complex valued with $f(x)=0$ for $x\notin (-a,a)$, we can approximate $f$ on $[-a,a]$ uniformly (https://en.wikipedia.org/wiki/Fej%C3%A9r%27s_theorem) with Fourrier partial sums, say, for large enough $k$,
$\sup_{x\in[-a,a]}|f_k(x) - f(x)| \le \epsilon$ and $f_k$ are sums of the form $\sum_{n=-k}^{k} c_n e^{i x u_n}$ for real numbers $c_n, u_n$ such that $f_k$ is $2a$-periodic. Then
by assumption of equality of characteristic functions $E[f_k(X)] = E[f_k(Y)]$ and
$$
E[f(X) - f(Y)]
= E[f(X) - f_k(X)] - E[f(Y)-f_k(Y)].
$$
Focusing on $\{|X|\le a\}$ and $\{|X|>a\}$ separately,
$$
|E[f(X) - f_k(X)]|
\le \epsilon P(|X|\le a) + P(|X|>a) \max_{|x|>a} |f_k(x)|.
$$
But since $f_k$ is periodic with $f_k(x) = f_k(x+2a)$ for all $x$ we have $\max_{|x|>a} |f_k(x)| = \max_{|x|\le a} |f_k(x)|$ and
$\max_{|x|\le a} |f_k(x)| \le \epsilon + \sup|f|$. In summary,
$$
|E[f(X)] - E[f_k(X)]| \le \epsilon + \epsilon(\epsilon+ \sup|f|).
$$
and the same holds for $Y$. Since this holds for any $\epsilon$ this proves that $E[f(X)]=E[f(Y)]$ for all bounded continuous $f$ supported in $[-a,a]$.

Let me mention here that the same proof also provides the result that

If $X_n$ is a sequence of random variable with charactersitic functions $\phi_n(t)\to \phi(t)$ for all real $t$ where $\phi(t)=E[e^{itX}]$ for some random variable $X$ then $E[f(X_n)]\to E[f(X)]$ for all bounded continuous with compact support, and consequently $X_n\to^d X$ in distribution.

The only new technicality required is to prove that "tightness" holds in the sense that for any $\epsilon$ we have $\sup_n P(|X|>a) \le \epsilon$ for some $a$. We can use the following classical argument that is given in many textbooks. First $\phi(t)$ is continuous at 0, $|1-E[\cos(itX)]\le|1-\phi(t)|\le \epsilon$ for any $t\in(-u,u)$ for some $u\in(0,\pi/8)$ (here $\pi/8$ is not meaningful but we want the sin function to be monotonous on $[-u,u]$. Second,
$$|\frac{1}{2u} \int_{-u}^u (1-\text{Re}\phi_n(t))dt - \frac{1}{2u} \int_{-u}^u (1-\text{Re}\phi_n(t))dt|\le \epsilon
$$
by dominated convergence with respect to the uniform measure on $[-u,u]$ where $\text{Re}$ denotes the real part.
$$
2\epsilon \ge \epsilon+\frac{1}{2u} \int_{-u}^u (1-\text{Re}\phi(t))dt
\ge 
\frac{1}{2u} \int_{-u}^u (1-\text{Re}\phi_n(t)) dt=
\frac{1}{2u} \int_{-u}^u E[1-\cos(tX_n)] dt
= E[1-\sin(uX_n)/(uX_n)]
$$
by Fubini's theorem, and thanks to $
E[1-\sin(uX_n)/(uX_n)]\ge E[1-1/|uX_n|]\ge \frac 12 P(|X_n|>1/u)$. This proves
$$\sup_n (|X_n|>1/u) \le 4 \epsilon.$$
From here, following the same strategy as before, let $f$ be continuous supported om $[-a,a]$ for some $a>|u|$. Using an approximation of $f$ uniformly by some trigonometric polynomial $f_k$ such that
$\sup_{x\in[-a,a]}|f_k(x) - f(x)| \le \epsilon$ (e.g., by Fejer's theorem) we find
$e_n = E[f_k(X)]-E[f_k(X_n)] \to 0$ as $n\to+\infty$ and
$$
E[f(X) - f(X_n)]
= e_n + E[f(X) - f_k(X)] - E[f(X_n)-f_k(X_n)].
$$
Focusing on $\{|X_n|\le u\}$ and $\{|X|>u\}$ separately,
$$
|E[f(X) - f_k(X)]|
\le \epsilon P(|X|\le u) + P(|X|>u) \sup |f_k|
\le \epsilon + 4\epsilon(\epsilon + \sup |f|).
$$
Since $e_n\to0$ this proves $E[f(X_n)]\to E[f(X)]$ for all continuous function with compact support hence $X_n\to^d X$.
A: *

*With $f_X,f_Y$ the probability distributions, if $f_X-f_Y$ is non-zero then for $T$ large enough you'll have that $$g(x)=\sum_k (f_X(x-2\pi kT)-f_Y(x-2\pi kT))$$ is a non-zero $2\pi T$-periodic distribution and a non-zero finite signed measure on $[0,2\pi T)$.


*The key is that for all $m\in \Bbb{Z}$ $$\int_0^{2\pi T}e^{ixm/T} g(x)dx=\int_\Bbb{R} e^{i xm/T}f_X(x)dx-\int_\Bbb{R} e^{i xm/T}f_Y(x)dx=0$$


*Therefore we also have  $\int_0^{2\pi T} (\frac{1+\cos(\frac{x-a}{T})}{2})^ng(x)dx\ne 0$ for all $a,n$.


*Take a $2\pi T$-periodic continuous function such that $\int_0^{2\pi T} \phi(x)g(x)dx\ne 0$, take $n,L$ large enough to approximate $\phi$ with $$\phi_{L,n}(x)=\frac{1}{C_n L} \sum_{l=0}^{L-1} \phi(\frac{2\pi T l}{L}) (\frac{1+\cos(\frac{x-\frac{2\pi T l}{L}}{T})}{2})^n$$ ($C_n=\int_0^{2\pi T} (\frac{1+\cos(\frac{x}{T})}{2})^ndx$)
Since $g(x)dx$ is a finite measure on $[0,2\pi T)$ you'll have that $\int_0^{2\pi T} \phi_{L,n}(x)g(x)dx\approx  \int_0^{2\pi T} \phi(x)g(x)dx\ne 0$, contradicting the point 3.
