Why MDS $X_k$ is measurable with respect to $\mathcal F_k$? Definition: $X_n$ is a martingale difference sequence (MDS) with respect to a filtration $\mathcal{F}_n$ if it is adapted, integrable, and $E[X_n\mid \mathcal{F}_{n-1}]=0$. 
Let $Y_n:=X_n+Y_{n-1}$. If $X_n$ is an MDS, does that mean $Y_n$ is a martingale?
I saw this question that claims that it does, but I did not understand the proof. It says: $Y_n=\sum\limits_{k=0}^{n} X_k$, so that $E[Y_n\mid \mathcal{F}_{n-1}]= Y_{n-1}+E[X_n\mid \mathcal{F}_{n-1}]=Y_{n-1}$.
I think they mean: $E[Y_n\mid \mathcal{F}_{n-1}] = E[X_n + Y_{n-1}\mid\mathcal{F}_{n-1}]  = 0 + E[Y_{n-1}\mid\mathcal{F}_{n-1}]$
But I don't get why $E[Y_{n-1}\mid\mathcal{F}_{n-1}]=Y_{n-1}$.
Using induction, assuming $E[Y_{n-1}\mid\mathcal{F}_{n-2}] = Y_{n-2}$, we need to prove $E[Y_{n}\mid\mathcal{F}_{n-1}] = Y_{n-1}$. We get that:
\begin{align}
E[Y_{n}\mid\mathcal{F}_{n-1}] &= E[Y_{n-1}\mid\mathcal{F}_{n-1}] \\
 &= E[X_{n-1} + Y_{n-2}\mid\mathcal{F}_{n-1}] \\
&= E[X_{n-1}\mid\mathcal{F}_{n-1}] + E[Y_{n-2}\mid\mathcal{F}_{n-1}] \\
&= E[X_{n-1}\mid\mathcal{F}_{n-1}] + E[E[Y_{n-1}\mid\mathcal{F}_{n-2}]\mid\mathcal{F}_{n-1}]\\
 &= E[X_{n-1}\mid\mathcal{F}_{n-1}] + E[Y_{n-1}\mid\mathcal{F}_{n-2}]\\
 &= E[X_{n-1}\mid\mathcal{F}_{n-1}] + Y_{n-2}
\end{align}
So it's enough to prove that $E[X_{n-1}\mid\mathcal{F}_{n-1}] = X_{n-1}$. Or that $X_n$ is measurable with respect to $\mathcal F_n$. How can this be proved?
Please explain without assuming any knowledge since I'm studying this alone with google so I may not be familiar with trivial things. Thank you very much!
 A: You have answered your question yourself but you did not realize it yet. You have said that the $X_n$ are adapted to the filtration $\mathcal F_n$. What does that mean? Well, that means that $X_n$ are $\mathcal F_n$-measurable.
In the other post, they should have written that $Y_n$ can be written as
$$\tag{$*$}Y_n=Y_0+ \sum_{k=1}^n X_k$$
Assuming $Y_0=0$ (or constant) then we can conclude that $Y_n$ is also $\mathcal F_n$-measurable since it is a function of $\mathcal F_n$-measurable random variables.
This observation leads to
$$\mathbb E[Y_{n-1}\mid \mathcal F_{n-1}] =Y_{n-1}$$
That is just a property of the conditional expectation, if a random variable is measurable with respect to the $\sigma$-algebra it is conditioned on then you get the random variable itself. This actually solves the problem.
Things can change if $Y_0$ is not constant though. 
About your approach with induction: it is actually correct what you did, ending up whether we have $X_n$ is $\mathcal F_n$-measurbale which is indeed the case from the definition. But did you check the base case? Do we have 
$$\mathbb E[Y_1 \mid \mathcal F_0 ] = Y_0?$$ 
What we do have is
$$\mathbb E[Y_1\mid \mathcal F_0 ] = \mathbb E[ X_1 + Y_0 \mid \mathcal F_0] = \mathbb E[Y_0 \mid \mathcal F_0]$$ 
It boils down to the question whether $\mathbb E[Y_0\mid \mathcal F_0] = Y_0$ holds. That does not hold in general, but the case where $Y_0$ is constant it does, that is also why I said that things might change without the assumption that $Y_0$ is constant. 
Here are two questions for you to think about. Can you think of another assumption on $Y_0$ which leads to $\mathbb E[Y_0 \mid \mathcal F_0 ] =Y_0$? If you have that assumption, can you  argue why $Y_n$ is (still) $\mathcal F_n$-measurbale from ($*$)? 
