# Does $f_{n}(x)=\frac{1}{\sqrt{2\pi n}} e^{-\frac{x^2}{2\pi n}}$ converge a.s. and/or in measure

Consider the measure space $$(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu)$$, where $$\mathcal{B}(\mathbb{R})$$ is the Borel sigma algebra and $$\mu$$ is Lebesgue measure Does the following function $$f_{n}(x)=\frac{1}{\sqrt{2\pi n}} e^{-\frac{x^2}{2\pi n}}$$ converge a.s. and/or in measure.

Thoughts I've done a few examples of a.s. and in measure convergence before this question however those examples where mainly with indiactor type functions, so I'm not 100% sure how to go about it with a function that looks like this.

I claim that $$f_n(x)\rightarrow 0$$ a.e. and in measure.

Note that $$\lim_{n\rightarrow \infty} f_n(x)=\lim_{n\rightarrow \infty} \frac{1}{\sqrt{2\pi n}} e^{-\frac{x^2}{2\pi n}} = 0$$ is clear for all $$x\in \mathbb{R}$$. Thus, $$f_n\rightarrow 0$$ everywhere, so automatically almost every where(a.e.).

Moreover, let $$\epsilon >0$$ be given. And let $$E_{n,\epsilon}=\{ x\in \mathbb{R} : \frac{1}{\sqrt{2\pi n}} e^{-\frac{x^2}{2\pi n}} \geq \epsilon \}=f_n^{-1}([\epsilon,\infty)).$$

Then observe that $$\exists N\in \mathbb{N}$$ such that $$N> \frac{1}{2\pi \epsilon^2}$$ .

Then note that, for any $$n\geq N$$, $$f_n(x)\leq \frac{1}{\sqrt{2\pi n}}<\epsilon \hspace{2cm}\forall x\in \mathbb{R}.$$

It means that $$f_n^{-1}([\epsilon,\infty))=\emptyset \hspace{1cm}\forall n\geq N.$$

Thus,

$$\mu(E_{n,\epsilon}) = 0 \hspace{3cm}\forall n\geq N.$$

So,

$$\lim_{n\rightarrow \infty} \mu(E_{n,\epsilon})= 0$$.

In conclusion, $$f_n$$ converges to $$0$$ in measure.