How to show a sequence of random variables is $\mathcal{A_{n}}$ measurable and bounded The question: Let $\{ X_{i} \}_{i=1}^{\infty}$ be independent uniformly integrable random variables with common mean $\mu = 0$. Then, $Y_{n} \equiv \sum_{i=1}^{n} X_{i}$ is a martingale wrt $\{ X_{n} \}$.
To show this, the three following characteristics must be shown.
i) That $Y_{n}$ is $\sigma ( X_1, ..., X_n )$-measurable $\forall n$.
ii) $\mathbb{E}[|Y_{n}|] < \infty$
iii) $\mathbb{E}[ Y_{n+1}|\{ X_{i} \}_{i=1}^{n}] = Y_{n}$
I can do (iii) without issue, obviously assuming (i)-(ii) hold. But how do I show (i) and (ii), or rather, what is sufficient?
For (i) for example, $X_{n}$ is $X_{n}$-measurable obviously, then as $Y_{n}$ is a function of only $X_{n}$, is that sufficient to say that $Y_{n}$ is $X_{n}$-measurable?
(ii) I know is a direct result from $X_{n}$ being uniformly integrable. Ie, they are tight, do not diverge to infinity, and tail end does not impact the expectation much. But would it be sufficient to say this?
 A: Your approach seems good.
Note that $Y_n$ is the combination of $X_i$ for $i \in \{1,\dots, n\}$, which are all $\sigma(X_1,\dots,X_n)$-measurable since $\forall i \in \{1,\dots, n\}, \quad \sigma(X_1,\dots,X_i) \in \sigma(X_1,\dots,X_n)$.
To check integrability, you should use the independence and the fact that uniform integrability implies $\sup_{i \leq n}\mathbb{E}[\vert X_i \vert]\lt\infty$.
A: For (i), if $i \leq n$, then $X_i$ is $\sigma(X_1,...,X_n)$-measurable. Hence, $Y_n$ is the sum of $\sigma(X_1,...,X_n)$-measurable random variables. This implies that $Y_n$ is $\sigma(X_1,...,X_n)$-measurable.
For (ii), uniform integrability of the sequence implies that each $X_n$ is integrable. In fact we have something even stronger, namely 
$$\sup_n E|X_n| = \sup_n \Big[ E\big(|X_n|\mathbf{1}_{|X_n > \alpha|} \big) + E\big(|X_n|\mathbf{1}_{|X_n| \leq \alpha}\big) \Big] \leq 1 + \alpha < \infty$$
for suitably large $\alpha$, by uniform integrability. (Note that we did not need to appeal to independence in this argument.)
At any rate, we now have $E|Y_n| \leq \sum_{i=1}^n E|X_n|<\infty$.
