I am reading about so-called "likelihood ratio martingales" in this handout.
The example given is as follows.
Let $(X_n : n \ge 1)$ be a sequence of iid random variables (say, on a probability space $(\Omega, \mathcal{F}, P)$) with common density $g$. Suppose that $f$ is another density with the property that whenever $g(x) = 0$, then $f(x) = 0$. Set $L_0 = 1$ and $$ L_n = \prod_{i=1}^{n} \frac{f(X_i)}{g(X_i)}, \quad n \ge 1. $$ Then (the claim is that) $L = (L_n: n \ge 0)$ is a martingale w.r.t. the filtration $\{\mathcal{F}_n\}$, where $\mathcal{F}_n = \sigma(X_1, ..., X_n)$.
As with many of these claims/proofs regarding martingales, it is always stated that the first two properties (i.e., (i) $L$ is $\{\mathcal{F}_n\}$-adapted; and (ii) $E[|L_n|] < \infty$, for all $n$) are either "immediate," or "trivial," or "obvious," or what-have-you; and this example is no exception. However, I don't see why this is the case! Let's give it a shot...
(i) $L$ is $\{\mathcal{F}_n\}$-adapted means that each $L_n$ is $\mathcal{F}_n$-measurable. So, since $\mathcal{F}_n = \sigma(X_1, ..., X_n)$, we need to check that, for each Borel set $B$, $L_n^{-1}(B) \in \sigma(X_1, ..., X_n)$. Ok... so, let $B$ be a Borel set. Then $$L_n^{-1}(B) = \{\omega : L_n(\omega) \in B\} = \left\{\omega : \frac{f(X_1(\omega))f(X_2(\omega)) \cdots f(X_n(\omega))}{g(X_1(\omega))g(X_2(\omega)) \cdots g(X_n(\omega))} \in B\right\}. $$ Why is it obvious that this set belongs to $\mathcal{F}_n$?
ii) $X_i$ are independent and $L_n\geq 0$, So
$$E|L_n|=E(L_n)=\prod_{i=1}^{n}E(\frac{f(X_i)}{g(X_i)}) =\prod_{i=1}^{n}\int \frac{f(x_i)}{g(x_i)}g(x_i) dx_i=1\leq +\infty$$
(i) Since $f$ and $g$ are density $$L_n=\frac{f(X_1)\times \cdots \times f(X_n)}{g(X_1)\times\cdots \times g(X_n)}=h(X_1,\cdots , X_n)$$ so It is a measurable function of $(X_1,\cdots , X_n)$ so $\sigma(L_n)\subset \sigma(X_1,\cdots , X_n)$.