Expected value (Wikipedia) is defined as follows:
In general, if $X$ is a random variable defined on a probability space $(\Omega ,\Sigma ,\operatorname {P} )$, then the expected value of $X$, denoted by $\operatorname {E} [X]$, is defined as the Lebesgue integral $$\operatorname {E} [X]=\int _{\Omega }X(\omega )\,d\operatorname {P} (\omega ).$$ Note that the integral is with respect to the probability measure.
However, characteristic function (Wikipedia) seems to define expected value in a different way.
$$\operatorname {E} \left[e^{itX}\right] = \int _{\mathbb {R} }e^{itx}\,dF_{X}(x)$$ where $F_X$ is the cumulative distribution function of $X,$ and the integral is of the Riemann-Stieljes kind.
Based on the definition of expected value given in the characteristic function, we should have $$\operatorname {E}(X) = \int _{\mathbb {R} }x\,dF_{X}(x).$$
I am wondering whether the two integrals above coincide, as the former is a Lebesgue integral (probability measure is the integrator) whereas the latter is a Riemann-Stieljes integral (CDF is the integrator).
Clearly measure and CDF are functions of different as their domains are different (one involves collection of sets, another involves element of set).
Let $\mu_X(A)=P(X^{-1}(A))$. Then $\mu_X$ is a probability measure on the Borel sigma algebra of $\mathbb R$ and the relation between $\mu_X$ and $F_X$ is given by $F_X(x)=\mu_X(-\infty,x]$. Now the defintion of $\mu_X$ gives $\int fd\mu =\int_{\Omega} f(X(\omega)) dP(\omega)$ because this holds when $f$ is an indicator function, hence when $f$ is simple function, hence when $f$ is any non-negative measurable function or a function integrable w.r.t. $\mu_X$. Thus $\int e^{itX}dP=\int e^{itx} d\mu_X(x)=\int e^{itx} dF_X(x)$. Note that the first integral here is over $\Omega$ and the other two are integrals over $\mathbb R$.