This is a proof-verification request.
Claim: Let $(X,\mathscr M,\mu)$ be a measure space. Let $f_n$ ($n\in\mathbb N$) and $f$ be measurable, integrable, real-valued functions such that $(f_n)_{n\in\mathbb N}$ converges to $f$ in $L^1$ at a rate $O(1/n^p)$, where $p>1$. Then, $f_n\to f$ almost everywhere.
Note: If no assumption is made on the rate of convergence, then the best one can establish is the existence of a $\textit{sub}$sequence $(f_{n_k})_{k\in\mathbb N}$ converging to $f$ almost everywhere (Corollary 2.32 in Folland, 1999).
Proof of the Claim: Suppose there exists $M>0$ such that $\int|f_n-f|\,\mathrm d\mu\leq M/n^p$ for all $n\in\mathbb N$. For each $\varepsilon>0$ and $n\in\mathbb N$, let $$E(n,\varepsilon)\equiv \big\{x\in X\,\big|\,|f_n(x)-f(x)|\geq\varepsilon\big\}.$$ Then, $$\frac {M}{n^p}\geq\int|f_n-f|\,\mathrm d\mu\geq \int_{E(n,\varepsilon)}|f_n-f|\,\mathrm d\mu\geq\varepsilon\,\mu(E(n,\varepsilon))$$ for each $n\in\mathbb N$, so that $$\mu(E(n,\varepsilon))\leq\frac{M}{n^p\varepsilon}.$$ Defining $$E(\varepsilon)\equiv\bigcap_{m=1}^{\infty}\bigcup_{n=m}^{\infty}E(n,\varepsilon),$$ one has that $$\mu(E(\varepsilon))\underset{\forall m\in\mathbb N}{\leq}\mu\left(\bigcup_{n=m}^{\infty}E(n,\varepsilon)\right)\leq\sum_{n=m}^{\infty}\mu(E(n,\varepsilon))\leq\frac{M}{\varepsilon}\sum_{n=m}^{\infty}\frac{1}{n^p}\to0\quad\text{as $m\to\infty$},$$
since the series $\sum_{n=1}^{\infty}1/n^{p}=\zeta(p)$ converges. Therefore, $\mu(E(\varepsilon))=0$ for any $\varepsilon>0$, which implies also that $$\mu\left(\bigcup_{q\in\mathbb Q\cap(0,\infty)}E(q)\right)=0.$$
Now, if $x\in X$ is such that $f_n(x)\not\to f(x)$, then there exists some $q>0$ such that $q\in\mathbb Q$ and for each $m\in\mathbb N$, there exists some $n\geq m$ so that $|f_n(x)-f(x)|\geq q$. That is, $x\in E(q)$. Hence, the set where pointwise convergence fails is a subset of $\bigcup_{q\in\mathbb Q\cap(0,\infty)}E(q)$, completing the proof. $\quad\blacksquare$
Best Answer
Yes, your proof is correct.
Moreover, instead of $\frac{1}{n^p}$ there could have been any function $f$, such that $\Sigma_{i = 1}^\infty f(n)$ converges.