# A question about the theorem of characteristic function

The statement is: Let $$X$$ be a real-valued random variable with characteristic function $$\phi_X(.)$$.Let $$Z=N(0,1)$$ be independent of $$X$$.For each $$\sigma>0$$ the random variable $$X_{\alpha}=X+\sigma Z$$ has a density $$f_{\alpha}$$ given by,
$$f_{\alpha}(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{itx}\phi_{X}(t)e^{\frac{-\sigma^2t^2}{2}}dt$$ proof: Fix $$\sigma>0$$ using the independence of $$X$$ and $$Z$$ we have
$$P(X_{\sigma}{\leq\alpha})=\int_{\mathbb{R}}F_Z(\frac{\alpha-x}{\sigma})d\mu_X(x)$$ $$\int_{\mathbb{R}}\int_{-\infty}^{\frac{\alpha-x}{\sigma}}\frac{1}{\sqrt{2\pi}}e^\frac{-a^2}{a}da d\mu_X(x)$$ I have marked the step where I am getting confused. What I have understood is that $$P(X_{\sigma}\leq\alpha)=P(X+\sigma Z\leq \alpha)$$ now I am confused how we got this equals to the integral given in the proof.

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This follows from independence and the Fubini-Tonelli theorem. To see this, let $$\alpha \in \mathbb{R}$$ and $$A = \{(u,v) \in \mathbb{R}\times\mathbb{R} : u + \sigma v \leq \alpha \}$$. Then,
\begin{align} P(X + \sigma Z \leq \alpha) &= P((X, Z) \in A)\\ &= \int_\mathbb{R} \int_\mathbb{R} 1_A(u, v)f_X(u)f_Z(v)dudv\\ &=\int_\mathbb{R} \left( \int_\mathbb{R} 1_A(u, v)f_Z(v)dv \right) f_X(u)du\\ &= \int_\mathbb{R} \left(\int_{-\infty}^{(\alpha - x)/\sigma}f_Z(v)dv \right) f_X(u)du\\ &= \int_\mathbb{R}F_Z\left(\frac{\alpha - x}{\sigma}\right)f_X(u)du \\ &= \int_\mathbb{R}F_Z\left(\frac{\alpha - x}{\sigma}\right)d\mu_X (x) \end{align} The next line then follows from the fact that $$Z\sim \mathcal{N}(0, 1)$$. Note that I've assumed $$X$$ to have a density, to make it clear where independence comes in.