I'm working on the following problem from an old real analysis qualifier at my university:
Let $(\Omega,\mathcal{A},\mu)$ be a measure space. Prove directly from the definition of convergence almost everywhere that if for all $n$, $\mu\{x\in\Omega\,:\,|f_n(x)|>1/n\}<n^{-3/2}$, then $f_n\to 0$ $\mu$-a.e.
I think I have a solution, but I don't use the suspicious exponent $-3/2$ anywhere in the proof.
Solution
Set $F_n=\{x\in\Omega\,:\,|f_n(x)|>1/n\}$. Then $$ \bigcap_{n=1}^{\infty}F_n=\{x\in\Omega\,:\,f_n(x)>1/n\text{ for all }n\in\mathbb{N}\}=\{x\in\Omega\,:\,f_n(x)\not\to 0\}, $$ so $$ \mu\left(\bigcap_{n=1}^{\infty}F_n\right)=\lim_{n\to\infty}\mu(F_n)=0. $$
I would like a second opinion on whether or not this is correct. Thanks in advance.
EDIT: I'd like to redeem myself. Does this solution seem to work?
Solution
Set $E_k=\{x\in \Omega\,:\,|f_n(x)|>1/n\}$. Fix $x\in\Omega$ with $f_n(x)\not\to 0$. Then there exists an $N\in\mathbb{N}$ and a $\varepsilon>0$ so that $|f_n(x)|>\varepsilon$ whenever $n\geq N$. Choosing $N'\in\mathbb{N}$ so that $\varepsilon>1/N'$ and $N'\geq N$ shows that $x\in E_k$ whenever $k\geq N'$; in particular, $x\in\bigcup_{k=j}^{\infty}E_{k}$ for each $j$. Set $F_j=\bigcup_{k=j}^{\infty}E_k$. What we have just shown is that $$ \{x\in\Omega\,:\,f_{n}(x)\not\to 0\}\subset\bigcap_{j=1}^{\infty}F_j. $$ Furthermore $$ \mu(F_j)=\mu\left(\bigcup_{k=j}^{\infty}E_k\right)\leq\sum_{k=j}^{\infty}\mu(E_k)\leq\sum_{k=1}^{\infty}\mu(E_k)\leq\sum_{k=1}^{\infty}\frac{1}{k^{3/2}}<\infty $$ for each $j$, so that $$ \mu\left(\bigcap_{j=1}^{\infty}F_j\right)=\lim_{j\to\infty}\mu(F_j)=\lim_{j\to\infty}\mu\left(\bigcup_{k=j}^{\infty}E_k\right)=0. $$ Comparing this with above yields the result.
No, it is wrong. In order for $f_n(x)$ not to go to $0$, it is not necessary for $x$ to be in the intersection of all the $F_n$, though it must be in infinitely many.