I am reading "Probability with Martingales" by David Williams. On page 97 the following theorem and proof is stated:
Theorem: Let $C$ be a bounded non-negative previsible process so that, for some $K\in[0,\infty)$, $|C_n(\omega)|\le K$ for every $n$ and every $\omega$. Let $X$ be a supermartingale [respectively martingale]. Then $C\bullet X$ is a supermartingale [martingale] null at $0$.
Proof: Write $Y$ for $C\bullet X$. Since $C_n$ is bounded non-negative and $F_{n-1}$-measurable, $E[Y_n-Y_{n-1}|F_{n-1}]=C_nE[X_n-X_{n-1}|F_{n-1}]\le0$, [resp. $=0$].
In the above $C\bullet X$ (and $Y_n$) is defined as $(C\bullet X)_n:=\sum_{k=1}^{n} C_k(X_k-X_{k-1})=Y_n$ and in particular $(C\bullet X)_0=0$ and $Y_n-Y_{n-1}=C_n(X_n-X_{n-1})$.
Now, in the proof, I understand how the author gets the first equality, but how is the last inequality [equality for martingale] established and why does this conclude the theorem? I.e., how is it that (for the case of supermartingale) $C_nE[X_n-X_{n-1}|F_{n-1}]\le0$, and how does this conclude that $C\bullet X$ is supermartingale?
I get that $X_{n-1}$ is $F_{n-1}$-measurable, so $E[X_{n-1}|F_{n-1}]=E[X_{n-1}]$, but is $X_n$ somehow also $F_{n-1}$-measurable, and if this is the case how do we clarify this?
And, as previously mentioned, how does $E[Y_n-Y_{n-1}|F_{n-1}]=C_nE[X_n-X_{n-1}|F_{n-1}]\le0$ in itself conclude that $E[Y_n | F_{n-1}]\le Y_{n-1}$?
You state that $\mathbb{E}[X_{n-1} \mid F_{n-1}] = \mathbb{E}[X_{n-1}]$, but that is not right. Since $X_{n-1}$ is $F_{n-1}$-measurable, in fact $\mathbb{E}[X_{n-1} \mid F_{n-1}] = X_{n-1}$. Since $X$ is a supermartingale, we also have $\mathbb{E}[X_n \mid F_{n-1}] \leq X_{n-1}$. So, using linearity of conditional expectation, we have $$\mathbb{E}[X_n-X_{n-1}\mid F_{n-1}] = \mathbb{E}[X_n\mid F_{n-1}]-\mathbb{E}[X_{n-1} \mid F_{n-1}] \leq X_{n-1}-X_{n-1} = 0.$$ Since $C_n$ is nonnegative, the same inequality holds if we put $C_n$ out front.
Now we have established that $\mathbb{E}[Y_n-Y_{n-1}\mid F_{n-1}] \leq 0$, and you ask why does that imply $Y$ is a supermartingale? It should be clear that $Y_n$ is $F_n$-measurable and integrable, so all that remains is to show $\mathbb{E}[Y_n \mid F_{n-1}] \leq Y_{n-1}$. Similarly to what we did above, this follows from linearity of conditional expectation and the fact that $\mathbb{E}[Y_{n-1}\mid F_{n-1}] = Y_{n-1}$: \begin{align*} \mathbb{E}[Y_n \mid F_{n-1}] &= \mathbb{E}[(Y_n-Y_{n-1})+Y_{n-1} \mid F_{n-1}]\\ & = \mathbb{E}[Y_n-Y_{n-1} \mid F_{n-1}]+\mathbb{E}[Y_{n-1}\mid F_{n-1}]\\ &\leq 0 + Y_{n-1}\\ &=Y_{n-1}. \end{align*}