Show that $ A \in \mathcal{F}_{m} $ and $\mathbb{E}[X_m \mathbb{1}_{A}]= \mathbb{E}[X_n \mathbb{1}_{A}]$ for $m<n$ then $(X_n)_{n \in \mathbb{N}}$ is a martingale.
Considering a sequence $(X_n)_{n \in \mathbb{N}}$ of real, integrable random variables, adapted to the filtration $\mathcal{(F_{n})_{n \in \mathbb{N}}}$
Actually I'm familiar with similar problems but this has stumped me. Any hints appreciated.
You can try to show integrability and adaptedness of $(X_n)$. Regarding the martingale property we know that the characterising property can be written as
\begin{align*} E[X_{n} | \mathcal{F}_m] = X_m, \ \forall \ m<n \end{align*}
So you just need to show that the above property holds. But now notice that the condition given by the exercise is precisely what you need! Indeed, by definition of conditional expectation we have that $E[X_n|\mathcal{F}_m]$ is a random variable $Z$ which is $\mathcal{F}_m-$measurable, integrable and satisfies the following property
\begin{align*} \int_A Z dP = \int_A X_n dP, \ \forall \ A \in \mathcal{F}_m \end{align*}
Now simply notice that $X_m$ fits the role of $Z$:
\begin{align*} E[X_m 1_A] = E[X_n 1_A], \ \forall A \in \mathcal{F}_m \Leftrightarrow \int_A X_m dP = \int_A X_n dP, \ \forall \ A \in \mathcal{F}_m \end{align*}