Suppose that $\mathbf{F}_{n} \rightarrow \mathbf{F}$ and the characteristic functions are dominated by integrable function. Show that $\mathbf{F}$ has a density that is the limit of the density of $\mathbf{ F}_{n} $
My attempt
$\phi_{n}(t) \rightarrow \phi(t)$, $\forall t \in \mathbb{R}$, that happens by Levy continuity theorem. $\phi_{n}(t)$ are dominated by integrable function then $|\phi_{n}(t)|\leq g$ where $\int g(x)dx<\infty$. Then $\mathbf{F}_{n}=\int f_{n}dx$ where $f_{n}=\frac{1}{2\pi}\int e^{-itx}\phi_{n}(t)dt$.
Then $$\mathbf{F}=\lim_{n \rightarrow \infty} f_{n}(x)=\lim_{n \rightarrow \infty} \frac{1}{2\pi}\int e^{-itx}\phi_{n}(t)\,dt=\frac{1}{2\pi}\int e^{-itx}\phi(t)\,dt$$
I don't see how $F(x)=\int f(x)dx$.
Any help is appreciated
You're missing an integral in your series of equations for $\ \mathbf{F}\ $. They should read \begin{align} \mathbf{F}(x)&=\lim_{n\rightarrow\infty} \mathbf{F}_n(x)\\ &= \lim_{n\rightarrow\infty}\int_{-\infty}^xf_n(y)dy\\ &= \lim_{n\rightarrow\infty}\int_ {-\infty}^x\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ity}\phi_n(t)dtdy\\ &\cdots \end{align} etc.