According to various sources (Wikipedia, Tao's blog), the strong law of large numbers--
Let $X$ be a r.v. with expectation $\mu := \Bbb{E}X$. Then if $X_1, X_2, X_3, ...$ are an iid sequence of copies of $X$, then $$\Bbb{P}( \lim_{n \to \infty} \overline{X_n} = \mu ) = 1,$$ where $\overline{X_n} = (X_1 + ... + X_n)/n$ is the sample average.
does not require the common assumption $\operatorname{Var}(X) < \infty$ (and still holds true if $\mu = +\infty$).
Most proofs of the strong law that I've seen in some way rely on finite moments, a particularly common one being $\Bbb{E}X^4 < \infty$.
But sources including the aforementioned Tao's blog frequently mention the complexity involved in proving the strong law without such assumptions, and the proofs I've come across seem a little abstruse.
Working from the starting point that SLLN holds for bounded r.v.'s, I came up with the following argument that SLLN holds for generic nonnegative r.v.'s without any assumptions on second moments, but I'm not 100% confident about the interchange of summation and limit towards the end, and I'd appreciate some feedback on whether or not this approach works [or can be modified to work]:
Proof attempt: Let $X \geq 0$ be a nonnegative r.v. with finite expectation $\mu = \Bbb{E}X < \infty$. (WLOG the underlying sample space, on $X$ and all its iid copies, is $\Omega = [0, 1]$, the probability measure $\Bbb{P}$ is just Lebesgue measure, and the events are the Lebesgue measurable sets.)
We can write $X = \sum_{k = 1}^\infty A_k$, where $A_i := X _{\{i-1 < X \leq i\}}$ for $i = 1, 2, 3, ...$ Clearly each of the $A_i, i \geq 1$, are bounded above by $i$, so they all have finite means as well. Let $\mu_i := \Bbb{E} A_i$, then $\sum_i \mu_i = \mu$ by linearity of expectation.
We take our sequence $\{X_n \}_{n \geq 1}$ of iid copies of $X$, and we get sequences $\{(A_i)_n \}$ of iid copies of $A_i$ for each $i \geq 1$, so that $X_n = \sum_i (A_i)_n$. and also for the sample means, $\overline{X_n} = \sum_i \overline{(A_i)}_n$. Since $A_i$ is bounded, we are guaranteed that for each $i$, the event $$E_i := \{ \omega: \lim_n \overline{(A_i)}_n(\omega) = \mu_i \}$$ has $\Bbb{P}(E_i) = 1$, by our assumption that SLLN holds for bounded r.v.'s. If $$E := \cap_i E_i = \{ \omega: \lim_n \overline{(A_i)}_n( \omega ) = \mu_i, \text{ for all } i \geq 1 \},$$ then $\Bbb{P}(E) = 1$ because $E$ is a countable intersection of events with probability $1$.
Now from here, we would like to be able to interchange limit and sum: \begin{align*} \lim_n \overline{X_n} &= \lim_n \sum_i \overline{(A_i)_n} \\ &= \sum_i ( \lim_n \overline{(A_i)_n} ) \\ &= \sum_i \mu_i \\ &= \mu \text{ on all of } E, \ \end{align*}
but it's not clear that any of the typical conditions (monotone convergence, dominated convergence) which allow us to interchange limits and sums hold here. So we choose $\epsilon > 0$ and appeal to Egorov's theorem to construct, for each $n \geq 1$, a subset $F_n \subset E$ with $\Bbb{P}(F_n) > 1 - \epsilon/2^n$, so that $$\sum_i \overline{(A_i)_n} \to \overline{X_n} \text{ uniformly on } F_n.$$ Let $F := \cap_n F_n$, then $\Bbb{P}(F) > 1 - \sum_{n \geq 1} \epsilon/2^n = 1 - \epsilon$ and $\sum_i \overline{(A_i)_n} \to \overline{X_n}$ uniformly on $F$ for all $n \geq 1$. So on $F$, we are justified in interchanging the sum and the limit: $$\lim_n \overline{X_n}(\omega) = \sum_{i \geq 1} \lim_n \overline{(A_i)_n}(\omega) = \sum_{i \geq 1} \mu_i = \mu, \forall \omega \in F,$$ which implies $\Bbb{P}(\lim_n \overline{X_n}(\omega) = \mu) \geq \Bbb{P}(F) > 1 - \epsilon$ for every $\epsilon > 0$. Therefore $\Bbb{P}(\lim_n \overline{X_n}(\omega) = \mu) = 1$. $\blacksquare$
Does this work? Was uniform convergence applied correctly in the final stages?