Prove Kolmogorov’s strong law of large numbers from Breiman ergodic theorem

ergodic-theorylaw-of-large-numbersprobability

I've heard that Kolmogorov's strong law of large numbers is a simple consequence of the Breiman ergodic theorem. However, I am having difficulty proving this for myself.

Kolmogorov's SLLN:
Assume that $X_1, X_2, \dots$ are independent RV on $(\Omega,\mathcal{B},P)$ with means $\mu_1, \mu_2, \dots$ and variances $\sigma_1^2, \sigma_2^2, \dots$ such that $\displaystyle\sum_{k=1}^\infty \dfrac{\sigma_k^2}{k^2} < \infty$. Then
\begin{aligned} \overline{X_n}-\overline{\mu_n}\stackrel{a.s.}{\rightarrow} 0. \end{aligned} where $\overline{X_n}=\frac{1}{n}\sum_{k=0}^n X_k$ and $\overline{\mu_n}=\frac{1}{n}\sum_{k=0}^n\mu_k$.

Breiman ergodic theorem: Let $T$ be a measure-preserving transformation on a probability space $(\Omega',\mathcal{B}',P')$ and $(g_i)$ a sequence of random variables on that space such that $E[\sup_n |g_n|] < \infty$, and the limit $\lim_n g_n=g$ exists almost surely, then
\begin{aligned} \frac{1}{n}\sum_{k=0}^n g_k(T^{k}(\omega))\rightarrow E[g] \quad \text{for almost all }\omega. \end{aligned}

I think the idea is to set $g_n=|\overline{X_n}-\overline{\mu_n}|$, $\Omega'=\Omega\times\Omega\times\ldots$ (the product space) and the $T$ the shift operator. However, I'm a bit confused with the details (is the shift operator measure-preserving on that product space?).

Also, how do you show that if $\displaystyle\sum_{k=1}^\infty \dfrac{\sigma_k^2}{k^2} < \infty$, then $E[\sup_n |g_n|] < \infty$? If $g_k=|X_k-\mu_k|$, then I can show this by using the inequality $E[|X|]\leq \sum_{k=0}^{\infty} P(|X|>k)$, but in the other case it doesn't seem to work…

Any help would be appreciated!

Best Answer

The assumption $g_k \to g$ in Breiman's theorem does not seem to have an analog in the SLLN presented. On the other hand, this SLLN follows from Kronecker's lemma which we quote next, by setting $b_k=k$ in [1]:

Kronecker's Lemma If $(y_n)_{n=1}^\infty$ is an infinite sequence of real numbers such that $\sum_{m=1}^\infty y_m$ exists and is finite, then
$$\lim_{n \to \infty}\frac1{n}\sum_{k=1}^n ky_k = 0 \,.$$

To prove the proposed SLLN, set $y_k=(X_k-\mu_k)/k$. Then $M_n=\sum_{k=1}^n y_k$ is a Martingale bounded in $L^2$, so it converges to a finite limit almost surely, see e.g. Prop. 1.7 in [3]. (Kolmogorov's original proof of this convergence used his eponymous inequality, see e.g. [2]). Now Kronecker's lemma can be applied.

[1] https://en.wikipedia.org/wiki/Kronecker%27s_lemma

[2] https://www.math.hkust.edu.hk/~makchen/MATH5411/Chap1Sec6.pdf

[3] http://www.math.stonybrook.edu/~bishop/classes/math627.S22/bloch_notes.pdf

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