Boundedness of characteristic function (using Fourier transform)

Question

Let $\mu$ be a probability measure on $\mathbb{R}^n$. Its Fourier transform is denoted $\mu'$ is a function on $\mathbb{R}^n$ is given by $$ \mu'(u)= \int e^{i\langle u,x\rangle} \mu(dx) $$ Now it says that it follows that $$ |\mu'|\leq \int |e^{i\langle u,x\rangle}| \mu(dx) = \int 1 \mu(dx) = 1 $$ now I understand the first steps, but can't understand why $\int 1 \mu(dx) = 1$ in the end. Writing it as characteristic functions I get it, because there $E(1)=1$ using the expected value. How is $\int 1 \mu(dx) = 1$ here? Appreciate your help.

Answer 1

$\mu$ is a probability measure and thus

$$\int 1 \mu(dx) = \mu(\mathbb{R}^n) = 1$$

Recall that $1$ is the indicator function on the set $\mathbb{R}^n$ and the integral of an indicator function $I_A$ is (by definition of the Lebesgue-integral) equal to the measure of $A$, i.e. $\mu(A)$.