Show that $X_t:=\mathbb{E}[Y|\mathcal{F_t}] $ is a martingale

Question

I have this exercise about martingales:

"Let $Y$ be a random variable with $\mathbb{E}(|Y|)<\infty$ and let $\mathbb{F}$ be a filtration as well as $X_t:=\mathbb{E}[Y|\mathcal{F_t}] $ for all $t \in I$. Show that $X$ is an $\mathbb{F}$-martingale."

Now, I'm new at martingales and would like to simply use the definition.

1. First of all, I should prove that $\mathbb{E}(\mathbb{E}(Y|\mathcal{F_t})) < \infty$ and I know that $\mathbb{E}(\mathbb{E}(Y|\mathcal{F_t}))=\mathbb{E}(Y|\mathcal{F_t})$ and I'd like to use tower law to say that $\mathbb{E}(Y|\mathcal{F_t})=\mathbb{E}(Y)$ but is that true? I know the tower law for two random variables and I was wondering if it stands also for a random variable and a $\sigma$-algebra. I believe it could work if $Y$ is $\mathcal{F_t}$-measurable, but don't see if it is the case.
2. Say that's proven, how can I show that $\mathbb{E}(\mathbb{E}(Y|\mathcal{F_t})|\mathcal{F_s})=\mathbb{E}(Y|\mathcal{F_t})$ for all $t>s$?

Answer 1

To see that $X_t$ is a martingale, first note that $X_t$, being a conditional expectation conditioned on $\mathcal{F}_t$, is $\mathcal{F}_t$-adapted. We then show it has finite expectation for all $t \in I$: $$\mathbb{E}(|X_t|) = \mathbb{E}(|\mathbb{E}(Y\, | \, \mathcal{F}_t)|) \leq \mathbb{E}(\mathbb{E}(|Y|\, | \, \mathcal{F}_t)) = \mathbb{E}(|Y|) < \infty$$

We then show the martingale property; for $t>s$: $$\mathbb{E}(X_t \, \, | \, \mathcal{F}_s) = \mathbb{E}(\mathbb{E}(Y\, | \, \mathcal{F}_t) \, \, | \, \mathcal{F}_s) = \mathbb{E}(Y\, | \, \mathcal{F}_s) = X_s$$

where the second equality is the tower property of conditional expectations.

Remarks

1. Note that the tower property holds for general $\sigma$-algebras, not just for $\sigma$-algebras generated by a single random variable.
2. The following is true: $\mathbb{E}(\mathbb{E}(Y\,|\, \mathcal{F}_t)) = \mathbb{E}(Y)$. In general, it is not true that $\mathbb{E}(Y\,|\, \mathcal{F}_t) = \mathbb{E}(Y)$.