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While reading Loeve's book on Probability Theory I had following doubt: in a basic course of probability I was taught that the moment generating function of a continuous random variable is computed as:

$E[e^{tX}] = \int_{-\infty}^{+\infty} e^{tx}f(x)dx$

where f(x) is the density function.

I want deduce this formula from the basic expression

$E[g(X)]=\int_{-\infty}^{+\infty}g \ dP_{X}$

which appears on the book. I was able to do so in the discrete case but can't figure out how to do it if g(X) is continuous.

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The density function is just the Radon derivative satisfying $$\int 1_A(x)dP_X(x)=P_X(A)=\int_A f(x)dx$$ Then when $g(x)$ is simple function we have $$\int g(x)dP_X(x)=\int g(x)f(x)dx$$ By monotone convergence and monotone convergence respectively, the above equation is also true when $g(x)$ is positive measurable function and bounded measurable function.

$e^{tx}$ is positive measurable function, so we have $$E[e^{tX}] = \int_{-\infty}^{+\infty} e^{tx}f(x)dx$$

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