Prove Kolmogorov's SLLN by martingale.

Question

Suppose $\xi_i$ are i.i.d. and $\mathbb E(|\xi_1|)\lt\infty$

Let $X_n=\sum_{i=1}^n\xi_i$ Then we have $\frac{X_n}{n}\to \mathbb E(|\xi_1|) $a.s.

In the proof of this theorem:

$$\frac{X_n}{n}=\frac{1}{n}\sum_{i=1}^n\mathbb E[\xi_i|X_n]$$ I don't know why this equality holds since I don't think $\xi_i\in\sigma(X_n)$

Answer 1

Indeed, $\xi_i$ has no reason to be $\sigma(X_n)$ measurable. However, we can use the following fact:

for each $i,j\in\{1,\dots,n\}$, $(\xi_i,X_n)=(\xi_j,X_n)$ in distribution.

Then notice that $X_n=\mathbb E[X_n\mid X_n]=\sum_{j=1}^n\mathbb E[\xi_j\mid \sigma(X_n)]$ (which is the formula in the OP up to the $1/n$ factor).

Answer 2

$$ X_n={\rm E}[X_n\mid X_n]=\sum_{i=1}^n {\rm E}[\xi_i\mid X_n] $$