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I am trying to prove the following fact:

Let $X$ be real-valued random variable and $\phi$ its characteristic function. Then for every real $a$ the following holds: $P(X=a) = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} e^{-ita}\phi(t)dt$.

I've reduced this identity to the following one:

$P(X=a) = \lim_{T \to \infty} \int_{\mathbb{R}} \frac{\sin T(x-a)}{T(x-a)}dP_X(x)$.

Not it sure now why it should hold. I would be grateful for your suggestions or ideas.

Thanks

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I think you should consult any decent textbook on Fourier analysis. –  Harald Hanche-Olsen Dec 15 '12 at 12:29
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1 Answer

up vote 1 down vote accepted

Observe that

$$\left|\frac{\sin(T \cdot (x-a))}{T \cdot (x-a)}\right| \leq \frac{1}{T \cdot (x-a)} \to 0 \qquad (T \to \infty)$$

for all $x \not= a$. For $x=a$ we have

$$\left|\frac{\sin(T \cdot (x-a))}{T \cdot (x-a)}\right| = 1 \to 1 \qquad (T \to \infty)$$

Since the integrand is bounded (by 1) and $\mathbb{P}_X$ is a finite measure, we can apply dominated convergence and obtain

$$\int_{\mathbb{R}} \frac{\sin(T \cdot (x-a))}{T \cdot (x-a)} \, d\mathbb{P}_X(x) \to \int_{\mathbb{R}} 1_{\{a\}}(x) \, d\mathbb{P}_X(x) = \mathbb{P}[X=a]$$

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Thanks a lot, saz! –  eugen1806 Dec 16 '12 at 9:09
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