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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.

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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.

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