$f_{n} \in L^{p}(X),$ such that $\lVert f_{n}-f_{n+1}\rVert_{p} \leq \frac{1}{n^2}$. Prove $f_{n}$ converges a.e.

Question

Here's the problem at hand:

Let $(X,\mu)$ be a measure space, let $f_{n} \in L^{p}(X)$ for each $n$ and some $p>1$, such that $\lVert f_{n}-f_{n+1}\rVert_{p} \leq \frac{1}{n^2}$. If $A_{n} = \left\lbrace x \in X : \lvert f_{n}(x)-f_{n+1}(x)\rvert \geq \frac{1}{n} \right\rbrace$, prove that:

1. $\mu(A_{n}) \leq \frac{1}{n^p};$
2. $\mu\mathopen{}\left(\overline{\lim}A_{n}\right)\mathclose{}=0;$
3. $f_{n}$ converges a.e.

I've done 1. and 2, but I can't finish 3. Here's the proof of 1. and 2:

1. follows from $\frac{1}{n^p} \mu(A_{n}) = \int_{A_{n}}\lvert f_{n}-f_{n+1}\rvert^p d\mu \leq \int_{X} \left\lvert f_{n}-f_{n+1}\right\rvert^p d\mu \leq \frac{1}{n^{2p}}$.

For 2, note that $\overline{\lim}A_{n} = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_{k} =: \bigcap_{n=1}^{\infty} B_{n}$. $B_{n}$ is a decreasing sequence, and $\mu(B_{1}) \leq \sum_{n=1}^{\infty} \frac{1}{n^p} < +\infty$, so by continuity from above, we get $ 0 \leq \mu\mathopen{}\left(\overline{\lim}A_{n}\right)\mathclose{} = \lim_{n\to \infty} \mu(B_{n}) \leq \lim_{n\to \infty} \sum_{k=n}^{\infty} \frac{1}{n^p} = 0$.

Now, I also note that $\lVert f_{m} - f_{n}\rVert_{p} \leq \sum_{k=n+1}^{m} \frac{1}{k^2} < \varepsilon$, for sufficiently large $m, n$, and arbitrary $\varepsilon$, so $f_{n}$ is Cauchy in $L^p$ so it's convergent in $L^p$; so now, as a corollary, there exists an $f$ such that $f_{n} \xrightarrow{\mu} f$, and so we have a subsequence $f_{n_{k}}$ which converges to $f$ a.e.

How do I prove that the whole sequence has to converge to $f$ a.e?

Answer 1

Alright, here's how I just figured it out (although I didn't use parts a) and b); I guess they were there to fool the problem-solver):

Define $\alpha = \frac{1-\frac{1}{p}}{2}$. Define $C_{n} = \lbrace x \in X : |f_{n}(x) - f_{n+1}(x)| \geq \frac{1}{n^{1+\alpha}}\rbrace$. Then, just as in 1, $\frac{1}{n^{2p}} \geq \mu(A_{n})\frac{1}{n^{p(1+\alpha)}}$; $2p-p-p\alpha = p-\frac{p-1}{2} = \frac{1+p}{2} > 1$, so we're good; $\mu(A_{n}) \leq \frac{1}{n^{\frac{1+p}{2}}}$.

For 2, proceed the same, and use that $\sum_{n=1}^{\infty} \frac{1}{n^{\frac{1+p}{2}}}$ converges, taking the limit of $\mu(D_{n})$, $D_{n} = \bigcup_{k=n}^{\infty}C_{k}$. So $\mu(\overline{\lim}C_{n})=0$.

Now, for 3, if $x \notin \overline{\lim}B_{n}$, after some $N$, we have $\forall n \geq N$ $|f_{n}(x) - f_{n+1}(x)| < \frac{1}{n^{1+\alpha}}$. Now, if $m>n$, and $\varepsilon>0$ is arbitrary,

$$|f_{m}(x) - f_{n}(x)| \leq \sum_{k=n+1}^{m} |f_{k}(x)-f_{k-1}(x)| < \sum_{k=n+1}^{m} \frac{1}{n^{1+\alpha}} < \varepsilon,$$

where the last two inequalities hold for large enough $m, n$. So the sequence $\{f_{n}(x)\}_{n=1}^{\infty}$ is Cauchy in $\mathbb{C}$, so it's convergent. This holds for every $x \notin \overline{\lim}C_{n}$, so it holds everywhere except for a null set, so $\{f_{n}(x)\}$ converges a.e.