Let $\Omega, \mu$ be a probability space. A measurable function $f: \Omega \rightarrow \mathbb{R}$ is called Gaussian if $$\mu (f^{-1}(A))=\frac{1}{\sigma \sqrt{2\pi }}\int_Ae^{-x^2/2\sigma ^2} dx$$ for some $\sigma>0$. I'm reading a proof that the space of Guassian functions is closed in $L^2(\Omega)$. The proof goes as follows:
If $f_n \in L^2(\Omega)$ is a sequence converging (in the $L^2$ norm) to $f$, then since $\sigma_n^2= ||f_n||_2^2$, we have $\sigma_n \rightarrow \sigma$. Therefore, for every $g(x)\in C_c(\mathbb{R})$, we have
$$\lim_n \frac{1}{\sigma_n \sqrt{2\pi }}\int e^{-x^2/2\sigma_n ^2} g(x) dx=\frac{1}{\sigma \sqrt{2\pi }}\int e^{-x^2/2\sigma ^2} g(x)dx$$ (I guess this is by Lebesgue's dominated convergence theorem?). Then the author says "This shows that the distribution of $f$ (i.e. the measure $A \mapsto \mu(f^{-1}(A))$) is the measure $\frac{1}{\sigma \sqrt{2\pi }}e^{-x^2/2\sigma ^2}dx$."
How does this last line follow? Thank you!
The equation
$$\lim_{n \to \infty} \frac{1}{\sigma_n \sqrt{2\pi}} \int e^{-x^2/2\sigma_n^2} g(x) \, dx = \frac{1}{\sigma \sqrt{2\pi}} \int e^{-x^2/2\sigma^2} g(x) \, dx$$
shows that $\mu(f_n^{-1}(\cdot))$ converges in distribution to a normal distribution with variance $\sigma^2$. On the other hand, by assumption, $f_n$ converges in $L^2$ to $f$ and this implies in particular convergence in distribution, i.e. $\mu(f_n^{-1}(\cdot)) \stackrel{d}{\to} \mu(f^{-1}(\cdot))$. Hence, $\mu(f^{-1}(\cdot)) \sim N(0,\sigma^2)$.