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I don't understand the argument of the solution:

Exercise:

Let $\nu$ be a finite measures on $(E,\mathcal{A})$ and let $\mathcal{F}_{\infty}:=\sigma(\bigcup_{k=1}^\infty \mathcal{F})$. Assume that for all $n\geq 0$ we have $\nu\ll \mathbb{P}$ and that there exists a non-negative, $\mathcal{F}_n$-measurable and $\mathbb{P}$-integrable function $M_n$ such that $\nu(A)=\mathbb{E}(M_n\mathbb{1}_{A})$ for every $A\in\mathcal{F}_n$. Prove that $(M_n)_{n\geq 0}$ is a martingale.

solution:

By hypothesis $M$ is adapted and integrable. For all $A\in \mathcal{F}_n$ we have that $\mathbb{1}_A$ is $\mathcal{F}_n$- and $\mathcal{F}_{n+1}$-measurable. Thus $\mathbb{E}(M_{n+1}\mathbb{1}_A)=\nu(A)=\mathbb{E}(M_{n}\mathbb{1}_A)$

I don't see why it is a martingale. I see that $\mathbb{E}(M_{n+1}\mathbb{1}_A)=\mathbb{E}(\mathbb{1}_A \mathbb{E}(M_{n+1}\mid \mathcal{F}_n))=\mathbb{E}(M_{n}\mathbb{1}_A)$

It says that $\mathbb{1}_A \mathbb{E}(M_{n+1}\mid \mathcal{F}_n)$ and $M_{n}\mathbb{1}_A$ have the same expectation.

But why can I conclude that $\mathbb{E}(M_{n+1}\mid \mathcal{F}_n)=M_{n}$?

Is it because $M_{n}=E(M_n\mid\mathcal{F}_n)$?

I don't quite understand the argument.

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Indeed you have $E(E(M_{n+1}\mid\mathcal{F}_n)1_A)=E(M_n1_A)$ for any $A\in\mathcal{F}_n$. There is a result from measure theory that says if $X$ and $Y$ are integrable random variables and $E(X1_A)=E(Y1_A)$ for any measurable $A$ then $X=Y$ almost surely. In this case $X$ is $E(M_{n+1}\mid\mathcal{F}_n)$ and $Y$ is $M_n$.