# Understanding characteristic functions in probability theory.

I am studying characteristic functions in probability theory and I am struggling to understand the following equality.

$$\int_{-\infty}^{\infty}e^{itX}dF_X(x)=\int_{-\infty}^{\infty}e^{itX}f_X(x)dx$$

Why is this transformation true?

Wikipedia states that $F_X$ is the cumulative distribution function of $X$, and the integral is of the Riemann–Stieltjes kind. But we haven't learned that yet. How do I have to understand this equity without using Riemann–Stieltjes?

What I also don't understand is what the true meaning of integrating by $dF(x)$ is.

Please explain that in terms of someone who has only visited introductions to Measure Theory, Lebesgue Integrals, and Probability Theory.

Thank you very much for your efforts!

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From the expression you have stated you may see that $dF_X(x) = f_X(x) dx$... –  Fabian Feb 7 '12 at 15:18
Where I come from, we introduced the notation $\int f dF$, where $F$ is a cumlative distribution function to be just a shortcut for $\int f dP$ if $P$ is the probability measure corresponding to $F$, without any mention of the Rieman-Stieltjes integral (Probably to not carry two letters around for basically the same thing). Maybe your course does that, too? –  Jens Feb 8 '12 at 10:57
@Jens Does this really mean $\int X dF_X=\int X dP_X$ if X is a realvalued measurable function where $F_X$ is the corresponding cumulative distribution function to the probability measure $P_X$ of the random variable $X$? –  Aufwind Feb 8 '12 at 13:32
As I stated, that was my definition for $dF$, so yeah. (Note though, that $\int XdP_X$ does not really make sense unless $X$ lives on its image, you'd usually use $\int X dP$ or $\int x dP_X(x)$. You've made the same mistake in the original question, compare your equation to the Wikipedia.) –  Jens Feb 8 '12 at 15:31
@Jens, sorry for the hassle, but I have to know for sure: I can treat the integrals as $\int XdP=\int xdP_X(x)=\int xdF(x)=\int xf_X(x)dx$? So you mean I have to correct $\int_{-\infty}^{\infty}e^{itX}dF_X(x)$ to $\int_{-\infty}^{\infty}e^{itx}dF_X(x)$? –  Aufwind Feb 8 '12 at 17:22
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In cases in which the probability distribution has a density function, one has $$\mathbb E(e^{itX}) = \int_{-\infty}^\infty e^{itx} f_X(x)\; dx = \int_{-\infty}^\infty e^{itx} \; dF(x).$$ For discrete probability distributions (where all the probability is accounted for by point-masses), the first integral makes no sense but the second is still valid.
The Riemann--Stieltjes integral is defined as the limit as the partition grows finer, of $$\sum f(x^*) \; \Delta F(x) = \sum f(x^*) (F(x+\Delta x) - F(x))$$ where $x^*$ is between $x$ and $x+\Delta x$.